{"id":49302,"date":"2022-04-04T15:42:56","date_gmt":"2022-04-04T10:12:56","guid":{"rendered":"https:\/\/ccssmathanswers.com\/?p=49302"},"modified":"2022-04-07T11:11:04","modified_gmt":"2022-04-07T05:41:04","slug":"big-ideas-math-geometry-answers-chapter-9","status":"publish","type":"post","link":"https:\/\/bigideasmathanswers.com\/big-ideas-math-geometry-answers-chapter-9\/","title":{"rendered":"Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry"},"content":{"rendered":"

Are you stressed to make your kid more intelligent and math proficient? Don’t worry at all as we have come up with the best solution that gives you relief from your stress. Here is the best preparation guide that aids your kids & makes them focused on the concepts of the respective Grade. Big Ideas Math Geometry Answers<\/a> Chapter 9 Right Triangles and Trigonometry<\/strong> is the one that we are discussing in the above lines. By using the BIM Geometry Ch 9 Solution Key, you can quickly gain the subject skills and score the highest marks in the exams. Fun & Activity-based learnings are easy & the simplest way to understand the concepts thoroughly and clarify your doubts within minimal time.<\/p>\n

Big Ideas Math Book Geometry Answer Key Chapter 9 Right Triangles and Trigonometry<\/h2>\n

Want to score more marks in Ch 9 Right Triangles and Trigonometry? Download the BIM Geometry Chapter 9 Right Triangles and Trigonometry Answer key Pdf for free from the below available links. After downloading the lesson-wise big ideas math book solution key for geometry ch 9,<\/strong> students need to begin their preparation efficiently and revise them frequently. Check out the links available here and tap on the respective lesson link of the high school Big Ideas Math Geometry ch 9 Solution Book to learn the questions covered in the practice tests, chapter tests, assessments, etc. of Right Triangles and Trigonometry concepts.<\/p>\n

Right Triangles and Trigonometry Maintaining Mathematical Proficiency<\/h3>\n

Simplify the expression.<\/p>\n

Question 1.
\n\u221a75<\/p>\n

Answer:
\nsquare root of 75 = 5,625.<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\n\u221a75.
\nsquare root of 75 = 75 x 75.
\n75 x 75 = 5,625.
\n\u221a75 = 5,625.<\/p>\n

Question 2.
\n\u221a270<\/p>\n

Answer:
\nsquare root of 270 = 72,900.<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\n\u221a270.
\nsquare root of 270 = 270 x 270.
\n270 x 270 = 72900.
\n\u221a270 = 72900.<\/p>\n

Question 3.
\n\u221a135<\/p>\n

Answer:
\nsquare root of 135 = 18225.<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\n\u221a135.
\nsquare root of 135 = 135 x 135.
\n135 x 135 = 18225.
\n\u221a135 = 18,225.<\/p>\n

Question 4.
\n\\(\\frac{2}{\\sqrt{7}}\\)<\/p>\n

Answer:
\n2\/49 = 0.04.<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\nsquare root of 7 = 7 x 7.
\n7 x 7 = 49.
\n\\(\\frac{2}{\\sqrt{7}}\\).
\n2\/49 = 0.04.<\/p>\n

Question 5.
\n\\(\\frac{5}{\\sqrt{2}}\\)<\/p>\n

Answer:
\n5\/4 = 1.25.<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\nsquare root of 2 = 2 x 2.
\n2 x 2 = 4.
\n\\(\\frac{5}{\\sqrt{2}}\\).
\n5\/4 = 1.25.<\/p>\n

Question 6.
\n\\(\\frac{12}{\\sqrt{6}}\\)<\/p>\n

Answer:
\n12\/36 = 0.33.<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\nsquare root of 6 = 6 x 6.
\n6 x 6 = 36.
\n\\(\\frac{12}{\\sqrt{6}}\\).
\n12\/36 = 0.33.<\/p>\n

Solve the proportion.<\/p>\n

Question 7.
\n\\(\\frac{x}{12}=\\frac{3}{4}\\)<\/p>\n

Answer:
\nx = 9.<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\n\\(\\frac{x}{12}=\\frac{3}{4}\\)
\nx\/12 = 3\/4.
\n4x = 12 x 3.
\n4x = 36.
\nx = 36\/4.
\nx = 9.<\/p>\n

Question 8.
\n\\(\\frac{x}{3}=\\frac{5}{2}\\)<\/p>\n

Answer:
\nx = 7.5.<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\n\\(\\frac{x}{3}=\\frac{5}{2}\\)
\nx\/3 = 5\/2.
\n2x = 5 x 3.
\n2x = 15.
\nx = 15\/2.
\nx = 7.5.<\/p>\n

Question 9.
\n\\(\\frac{4}{x}=\\frac{7}{56}\\)<\/p>\n

Answer:
\nx = 32.<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\n\\(\\frac{4}{x}=\\frac{7}{56}\\)
\n4\/x = 7\/56.
\n7x = 56 x 4.
\n7x = 224.
\nx = 224\/7.
\nx = 32.<\/p>\n

Question 10.
\n\\(\\frac{10}{23}=\\frac{4}{x}\\)<\/p>\n

Answer:
\nx = 9.2.<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\n\\(\\frac{10}{23}=\\frac{4}{x}\\)
\nx\/4 = 10\/23.
\n10x = 23 x 4.
\n10x = 92.
\nx = 92\/10.
\nx = 9.2.<\/p>\n

Question 11.
\n\\(\\frac{x+1}{2}=\\frac{21}{14}\\)<\/p>\n

Answer:
\nx = 135.<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\n\\(\\frac{x + 1}{2}=\\frac{21}{14}\\)
\nx+12 x 2 = 21×14.
\n2x + 24= 294 .
\n2x = 294 – 24.
\n2x = 270.
\nx = 270\/2.
\nx = 135.<\/p>\n

Question 12.
\n\\(\\frac{9}{3 x-15}=\\frac{3}{12}\\)<\/p>\n

Answer:
\nx = 6.33.<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\n\\(\\frac{9}{3 x-15}=\\frac{3}{12}\\)
\n27x – 135 = 3×12.
\n27x = 36 + 135.
\n27x = 171.
\nx = 171\/27.
\nx = 6.33.<\/p>\n

Question 13.
\nABSTRACT REASONING<\/strong>
\nThe Product Property of Square Roots allows you to simplify the square root of a product. Are you able to simplify the square root of a sum? of a diffrence? Explain.<\/p>\n

Answer:
\nYes, I am able to simplify the square root of a sum.<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\nThe product property of square roots allows you to simplify the square root of a product.
\n\u221a3 + 1 = \u221a4.
\n\u221a4 = 4 x 4.
\n16.
\n\u221a3 – 1 = \u221a2.
\n\u221a2 = 2 x 2.
\n4.<\/p>\n

Right Triangles and Trigonometry Mathematical practices<\/h3>\n

Monitoring progress<\/strong><\/p>\n

Question 1.
\nUse dynamic geometry software to construct a right triangle with acute angle measures of 30\u00b0 and 60\u00b0 in standard position. What are the exact coordinates of its vertices?<\/p>\n

Answer:<\/p>\n

Question 2.
\nUse dynamic geometry software to construct a right triangle with acute angle measures of 20\u00b0 and 70\u00b0 in standard position. What are the approximate coordinates of its vertices?
\nAnswer:<\/p>\n

9.1 The Pythagorean Theorem<\/h3>\n

Exploration 1<\/strong><\/p>\n

Proving the Pythagorean Theorem without Words<\/p>\n

Work with a partner.<\/p>\n

\"Big<\/p>\n

a. Draw and cut out a right triangle with legs a and b, and hypotenuse c.<\/p>\n

Answer:<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\nproving the Pythagorean theorem without words.
\na2<\/sup> + b2<\/sup> = c2.
\n<\/sup>\"Bid-Ideas-Math-Book-Geometry-Answer-Key-Chapter-9-Right<\/p>\n

b. Make three copies of your right triangle. Arrange all tour triangles to form a large square, as shown.<\/p>\n

Answer:
\na2<\/sup> + b2<\/sup> = c2.<\/sup><\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\nmake three copies of your right triangle.
\na2<\/sup> + b2<\/sup> = c2.<\/sup><\/p>\n

c. Find the area of the large square in terms of a, b, and c by summing the areas of the triangles and the small square.<\/p>\n

Answer:
\nThe area of the large square = a2<\/sup> x b2.<\/sup><\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\nthe area of the square = l x b.
\nwhere l = length, and b = breadth.
\nthe area of the square = a x b.
\narea = a2<\/sup> x b2.<\/sup><\/p>\n

d. Copy the large square. Divide it into two smaller squares and two equally-sized rectangles, as shown.<\/p>\n

Answer:<\/p>\n

e. Find the area of the large square in terms of a and b by summing the areas of the rectangles and the smaller squares.
\nAnswer:<\/p>\n

f. Compare your answers to parts (c) and (e). Explain how this proves the Pythagorean Theorem.<\/p>\n

Answer:
\na2<\/sup> + b2<\/sup> = c2.<\/sup><\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\nThe length of the a and b is equal to the hypotenuse.
\na2<\/sup> + b2<\/sup> = c2.
\nwhere a = one side and b = one side.<\/sup><\/p>\n

Exploration 2<\/strong><\/p>\n

Proving the Pythagorean Theorem<\/p>\n

Work with a partner:<\/p>\n

a. Draw a right triangle with legs a and b, and hypotenuse c, as shown. Draw the altitude from C to \\(\\overline{A B}\\) Label the lengths, as shown.
\n\"Big<\/p>\n

Answer:
\na2<\/sup> + b2<\/sup> = c2.<\/sup><\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\na2<\/sup> + b2<\/sup> = c2.<\/sup><\/p>\n

\"Bid-Ideas-Math-Book-Geometry-Answer-Key-Chapter-9-Right<\/p>\n

b. Explain why \u2206ABC, \u2206ACD, and \u2206CBD are similar.<\/p>\n

Answer:
\nIn a right-angle triangle all the angle are equal.<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\n\u2206ABC, \u2206ACD, and \u2206CBD are similar.
\nIn a right-angle triangle all the angles are equal.<\/p>\n

REASONING ABSTRACTLY<\/strong>
\nTo be proficient in math, you need to know and flexibly use different properties of operations and objects.
\nAnswer:<\/p>\n

c. Write a two-column proof using the similar triangles in part (b) to prove that a2<\/sup> + b2<\/sup> = c2<\/sup><\/p>\n

Answer:
\na2<\/sup> + b2<\/sup> = c2<\/sup><\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\nIn the pythagorean theorem.
\na2<\/sup> + b2<\/sup> = c2
\nthe length of the hypotenuse ie equal to the two side lengths.
\n<\/sup>a2<\/sup> + b2<\/sup> = c2<\/sup><\/p>\n

Communicate Your Answer<\/p>\n

Question 3.
\nHow can you prove the Pythagorean Theorem?<\/p>\n

Answer:
\na2<\/sup> + b2<\/sup> = c2<\/sup><\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\nIn the pythagorean theorem,
\nthe length of the hypotenuse is equal to the length of the other two sides.
\nhypotenuse = c.
\nlength = a.
\nthe breadth = b.
\na2<\/sup> + b2<\/sup> = c2<\/sup><\/p>\n

Question 4.
\nUse the Internet or sonic other resource to find a way to prove the Pythagorean Theorem that is different from Explorations 1 and 2.
\nAnswer:<\/p>\n

Lesson 9.1 The Pythagorean Theorem<\/h3>\n

Monitoring Progress<\/strong><\/p>\n

Find the value of x. Then tell whether the side lengths form a Pythagorean triple.<\/p>\n

Question 1.
\n\"Big<\/p>\n

Answer:
\nx = \u221a52.<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\nthe side lengths are 6 and 4.
\na2<\/sup> + b2<\/sup> = c2
\n6 x 6 + 4 x 4 = c2
\n36 + 16 = c2
\n52 = c2.
\nc = \u221a52.<\/sup><\/p>\n

Question 2.
\n\"Big
\nAnswer:
\nx = 4.<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\nthe side lengths are 3 and 5.
\nx<\/span>2 + 3 x 3 = 5 x 5.
\nx2 + 9 = 25.
\nx2 = 25 – 9.
\nx2 = 16.
\n<\/sup>x = 4.<\/p>\n

Question 3.
\nAn anemometer is a device used to measure wind speed. The anemometer shown is attached to the top of a pole. Support wires are attached to the pole 5 feet above the ground. Each support wire is 6 feet long. How far from the base of the pole is each wire attached to the ground?
\n\"Big<\/p>\n

Answer:
\nx = \u221a11.<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\nAn anemometer is a device used to measure wind speed.
\nsupport wires are attached to the pole 5 feet above the ground.
\nEach support wire is 6 feet long.
\nd<\/span>2 + 6\u00a0 x 6 = 5 x 5.
\nd2 + 36 = 25.
\nd2 = 25 – 36.
\nd2 = 11.
\nd<\/sup> = \u221a11.<\/p>\n

Tell whether the triangle is a right triangle.<\/p>\n

Question 4.
\n\"Big<\/p>\n

Answer:
\nYes the triangle is a right triangle.<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\nthe hypotenuse = 3 \u221a34.
\none side = 15.
\nthe other side = 9.
\nso the triangle is a right triangle.<\/p>\n

Question 5.
\n\"Big<\/p>\n

Answer:
\nYes, the triangle is a actute triangle.<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\nthe hypotenuse = 22.
\none side = 26.
\nthe other side = 14.
\nso the triangle is a acute triangle.<\/p>\n

Question 6.
\nVerify that segments with lengths of 3, 4, and 6 form a triangle. Is the triangle acute, right, or obtuse?<\/p>\n

Answer:
\nYes the lengths of the triangle form a acute triangle.<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\nthe side lenths of 3, 4, and 6 form a triangle.
\n6 x 6 = 3 x 3 + 4 x 4.
\n36 = 9 + 16.
\n36 = 25.
\nso the length of the triangle forms a acute triangle.<\/p>\n

\"Bid-Ideas-Math-Book-Geometry-Answer-Key-Chapter-9-Right<\/p>\n

Question 7.
\nVerify that segments with lengths of 2, 1, 2, 8, and 3.5 form a triangle. Is the triangle acute, right, or obtuse?
\nAnswer:<\/p>\n

Exercise 9.1 The Pythagorean Theorem<\/h3>\n

Vocabulary and Core Concept Check<\/p>\n

Question 1.
\nVOCABULARY<\/strong>
\nWhat is a Pythagorean triple?
\nAnswer:
\n\"Big<\/p>\n

Question 2.
\nDIFFERENT WORDS, SAME QUESTION<\/strong>
\nWhich is different? Find \u201cboth\u201d answers.
\n\"Big
\nFind the length of the longest side.<\/p>\n

Answer:
\nThe length of the longest side = 5.<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\nthe side lengths are 3 and 4.
\nin the pythagorean threoem,
\nthe longest side is equal to the side lengths.
\nX = 4 x 4 + 3 x 3.
\nX x X = 16 + 9.
\nX x X = 25.
\nX = 5.<\/p>\n

Find the length of the hypotenuse<\/p>\n

Answer:
\nThe length of the hypotenuse = 5.<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\nthe side lengths are 3 and 4.
\nin the Pythagorean theorem,
\nthe longest side is equal to the side lengths.
\nX = 4 x 4 + 3 x 3.
\nX x X = 16 + 9.
\nX x X = 25.
\nX = 5.<\/p>\n

Find the length of the longest leg.<\/p>\n

Answer:
\nThe length of the longest leg = 5.<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\nthe side lengths are 3 and 4.
\nin the pythagorean threoem,
\nthe longest side is equal to the side lengths.
\nX = 4 x 4 + 3 x 3.
\nX x X = 16 + 9.
\nX x X = 25.
\nX = 5.<\/p>\n

Find the length of the side opposite the right angle.<\/p>\n

Answer:
\nThe length of the side opposite to the right angle = 5.<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\nthe side lengths are 3 and 4.
\nin the pythagorean threoem,
\nthe longest side is equal to the side lengths.
\nX = 4 x 4 + 3 x 3.
\nX x X = 16 + 9.
\nX x X = 25.
\nX = 5.<\/p>\n

Monitoring progress and Modeling with Mathematics<\/p>\n

In Exercises 3-6, find the value of x. Then tell whether the side lengths form a Pythagorean triple.<\/p>\n

Question 3.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 4.
\n\"Big<\/p>\n

Answer:
\nThe length of the x = 34.<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\nthe side lengths are 30 and 16.
\nin the pythagorean threoem,
\nthe longest side is equal to the side lengths.
\nX = 16 x 16 + 30 x 30.
\nX x X = 256 + 900.
\nX x X = 1156.
\nX = 34.<\/p>\n

Question 5.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 6.
\n\"Big<\/p>\n

Answer:
\nThe length of the x = 7.2.<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\nthe side lengths are 6 and 4.
\nin the pythagorean threoem,
\nthe longest side is equal to the side lengths.
\nX = 4 x 4 + 6 x 6.
\nX x X = 16 + 36.
\nX x X = 52.
\nX = 7.2.<\/p>\n

In Exercises 7 – 10, find the value of x. Then tell whether the side lengths form a Pythagorean triple.<\/p>\n

Question 7.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 8.
\n\"Big<\/p>\n

Answer:
\nThe length of the X = 25.6.<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\nthe side lengths are 24 and 9.
\nin the pythagorean threoem,
\nthe longest side is equal to the side lengths.
\nX = 24 x 24 + 9 x 9.
\nX x X = 576 + 81.
\nX x X = 657.
\nX = 25.6.<\/p>\n

Question 9.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 10.
\n\"Big<\/p>\n

Answer:
\nThe length of the x = 11.4.<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\nthe side lengths are 7 and 9.
\nin the pythagorean threoem,
\nthe longest side is equal to the side lengths.
\nX = 7 x 7 + 9 x 9.
\nX x X = 49 + 81.
\nX x X = 130.
\nX = 11.4.<\/p>\n

ERROR ANALYSIS<\/strong>
\nIn Exercises 11 and 12, describe and correct the error in using the Pythagorean Theorem (Theorem 9.1).<\/p>\n

Question 11.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 12.
\n\"Big<\/p>\n

Answer:
\nx = 24.<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\nthe side lengths are 26 and 10.
\n26 x 26 = a x a + 10 x 10.
\n676 = a x a + 100.
\n676 – 100 = a x a.
\n576 = a x a.
\na = 24.
\nx = 24.<\/p>\n

Question 13.
\nMODELING WITH MATHEMATICS<\/strong>
\nThe fire escape forms a right triangle, as shown. Use the Pythagorean Theorem (Theorem 9. 1) to approximate the distance between the two platforms. (See Example 3.)
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 14.
\nMODELING WITH MATHEMATICS<\/strong>
\nThe backboard of the basketball hoop forms a right triangle with the supporting rods, as shown. Use the Pythagorean Theorem (Theorem 9.1) to approximate the distance between the rods where the meet the backboard.
\n\"Big<\/p>\n

Answer:
\nx = 9.1.<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\nthe side lengths are 13.4 and 9.8.
\n13.4 x 13.4 = X x X + 9.8 x 9.8.
\n179.56 = X x X + 96.04.
\n179.56 – 96.04 = X x X.
\n83.52 = X x X.
\nX = 9.1.
\nIn Exercises 15 – 20, tell whether the triangle is a right triangle.<\/p>\n

Question 15.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 16.
\n\"Big<\/p>\n

Answer:
\nNo, the triangle is not a right triangle.<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\nthe side lengths are 23 and 11.4.
\nthe hypotenuse = 21.2.
\n21.2 x 21.2 = 23 x 23 + 11.4 x 11.4.
\n449.44 = 529 + 129.96.
\n449.44 = 658.96.
\n449 is not equal to 658.96.
\nso the triangle is not a right triangle.<\/p>\n

Question 17.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 18.
\n\"Big<\/p>\n

Answer:
\nNo, the triangle is not a right triangle.<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\nthe side lengths are 5 and 1.
\nthe hypotenuse = \u221a26.
\n26 x 26 = 5 x 5 + 1 x 1.
\n676 = 25 + 1.
\n676 = 26.
\n676 is not equal to 26.
\nso the triangle is not a right triangle.<\/p>\n

Question 19.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 20.
\n\"Big<\/p>\n

Answer:
\nYes, the triangle forms a right triangle.<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\nthe side lengths are 80 and 39.
\nthe hypotenuse = 89.
\n89 x 89 = 80 x 80 + 39 x 39.
\n7921 = 6400 + 1521.
\n7921 = 7921.
\n7921 is equal to 7921.
\nso the triangle forms a right triangle.<\/p>\n

In Exercises 21 – 28, verify that the segment lengths form a triangle. Is the triangle acute, right, or obtuse?<\/p>\n

Question 21.
\n10, 11, and 14
\nAnswer:
\n\"Big<\/p>\n

Question 22.
\n6, 8, and 10<\/p>\n

Answer:
\nYes, the triangle is forming a right triangle.<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\nthe side lengths are 8 and 6.
\nthe hypotenuse = 10.
\n10 x 10 = 8 x 8 + 6 x 6.
\n100 = 64 + 36.
\n100 = 100.
\n100 is\u00a0 equal to 100.
\nso the triangle is forming a right triangle.<\/p>\n

Question 23.
\n12, 16, and 20
\nAnswer:
\n\"Big<\/p>\n

Question 24.
\n15, 20, and 36<\/p>\n

Answer:
\nYes, the triangle is obtuse triangle.<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\nthe side lengths are 15 and 20.
\nthe hypotenuse = 36.
\n36 x 36 = 20 x 20 + 15 x 15.
\n1296 = 400 + 225.
\n1296 > 625.
\n1296 is greater than 625.
\nso the triangle is not a obtuse triangle.<\/p>\n

Question 25.
\n5.3, 6.7, and 7.8
\nAnswer:
\n\"Big<\/p>\n

Question 26.
\n4.1, 8.2, and 12.2<\/p>\n

Answer:
\nNo, the triangle is obtuse triangle.<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\nthe side lengths are 4.1 and 8.2.
\nthe hypotenuse = 12.2.
\n12.2 x 12.2 = 4.1 x 4.1 + 8.2 x 8.2`.
\n148.84 = 16.81 + 67.24.
\n148.84 > 84.05.
\n148.84 is greater than 84.05.
\nso the triangle is obtuse triangle.<\/p>\n

Question 27.
\n24, 30, and 6\u221a43
\nAnswer:
\n\"Big<\/p>\n

Question 28.
\n10, 15 and 5\u221a13<\/p>\n

Answer:
\nYes, the triangle is an acute triangle.<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\nthe side lengths are 10 and 5\u221a13.
\nthe hypotenuse = 15.
\n15 x 15 = 10 x 10 + 5\u221a13 x 5\u221a13.
\n225 = 100 + 34.81.
\n225 < 134.81.
\n225 is less than 134.81.
\nso the triangle is acute triangle.<\/p>\n

Question 29.
\nMODELING WITH MATHEMATICS<\/strong>
\nIn baseball, the lengths of the paths between consecutive bases are 90 feet, and the paths form right angles. The player on first base tries to steal second base. How far does the ball need to travel from home plate to second base to get the player out?
\nAnswer:
\n\"Big<\/p>\n

Question 30.
\nREASONING<\/strong>
\nYou are making a canvas frame for a painting using stretcher bars. The rectangular painting will be 10 inches long and 8 inches wide. Using a ruler, how can you be certain that the corners of the frame are 90\u00b0
\n\"Big<\/p>\n

Answer:
\nx = 12.8.<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\nthe side lengths are 10 and 8.
\nthe hypotenuse = x.
\nX x X = 10 x 10 + 8 x 8.
\nX = 100 + 64.
\nX = 12.8.
\nIn Exercises 31 – 34, find the area of the isosceles triangle.<\/p>\n

Question 31.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 32.
\n\"Big<\/p>\n

Answer:
\nThe area of the Isosceles triangle = 12 ft.<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\nbase = 32 ft.
\nhypotenuse = 20ft.
\na2<\/sup> + b2<\/sup> = c2
\n<\/sup>h x h + 16 x 16 = 20 x 20.
\nh x h + 256 = 400.
\nh x h = 400 – 256.
\nh x h = 144.
\nh = 12 ft.
\nso the area of the isosceles triangle = 12 ft.<\/p>\n

Question 33.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 34.
\n\"Big<\/p>\n

Answer:
\nThe area of the Isosceles triangle = 48 m.<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\nbase = 28 m.
\nhypotenuse = 50 m.
\na2<\/sup> + b2<\/sup> = c2
\n<\/sup>h x h + 14 x 14 = 50 x 50.
\nh x h + 196 = 2500.
\nh x h = 2500 – 196.
\nh x h = 2304.
\nh = 48 m.
\nso the area of the isosceles triangle = 48 m.<\/p>\n

Question 35.
\nANALYZING RELATIONSHIPS<\/strong>
\nJustify the Distance Formula using the Pythagorean Theorem (Thin. 9. 1).
\nAnswer:
\n\"Big<\/p>\n

Question 36.
\nHOW DO YOU SEE IT?<\/strong>
\nHow do you know \u2220C is a right angle without using the Pythagorean Theorem (Theorem 9.1) ?
\n\"Big<\/p>\n

Answer:
\nYes, the triangle is forming a right triangle.<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\nthe side lengths are 8 and 6.
\nthe hypotenuse = 10.
\n10 x 10 = 8 x 8 + 6 x 6.
\n100 = 64 + 36.
\n100 = 100.
\n100 is\u00a0 equal to 100.
\nso the triangle is forming a right triangle.<\/p>\n

Question 37.
\nPROBLEM SOLVING<\/strong>
\nYou are making a kite and need to figure out how much binding to buy. You need the binding for the perimeter of the kite. The binding Comes in packages of two yards. How many packages should you buy?
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 38.
\nPROVING A THEOREM<\/strong>
\nUse the Pythagorean Theorem (Theorem 9. 1) to prove the Hypotenuse-Leg (HL) Congruence Theorem (Theorem 5.9).
\n\"Big
\nAnswer:<\/p>\n

Question 39.
\nPROVING A THEOREM<\/strong>
\nProve the Converse of the Pythagorean Theorem (Theorem 9.2). (Hint: Draw \u2206ABC with side lengths a, b, and c, where c is the length of the longest side. Then draw a right triangle with side lengths a, b, and x, where x is the length of the hypotenuse. Compare lengths c and x.)
\nAnswer:
\n\"Big<\/p>\n

Question 40.
\nTHOUGHT PROVOKING<\/strong>
\nConsider two integers m and n. where m > n. Do the following expressions produce a Pythagorean triple? If yes, prove your answer. If no, give a counterexample.
\n2mn, m2<\/sup> – n2<\/sup>, m2<\/sup> + n2<\/sup><\/p>\n

Answer:<\/p>\n

Question 41.
\nMAKING AN ARGUMENT<\/strong>
\nYour friend claims 72 and 75 Cannot be part of a pythagorean triple because 722<\/sup> + 752<\/sup> does not equal a positive integer squared. Is your friend correct? Explain your reasoning.
\nAnswer:
\n\"Big<\/p>\n

Question 42.
\nPROVING A THEOREM<\/strong>
\nCopy and complete the proof of the pythagorean Inequalities Theorem (Theorem 9.3) when c2<\/sup> < a2<\/sup> + b2<\/sup>.
\nGiven In \u2206ABC, c2<\/sup> < a2<\/sup> + b2<\/sup> where c is the length
\nof the longest side.
\n\u2206PQR has side lengths a, b, and x, where x is the length of the hypotenuse, and \u2220R is a right angle.
\nProve \u2206ABC is an acute triangle.
\n\"Big<\/p>\n\n\n\n\n\n\n\n\n\n\n\n\n
Statements<\/td>\nReasons<\/td>\n<\/tr>\n
1. In \u2206ABC, C2 < (12 + h2, where c is the length of the longest side. \u2206PQR has side lengths a, b, and x, where x is the length of the hypotenuse, and \u2220R is a right angle.<\/td>\n1. _____________________________<\/td>\n<\/tr>\n
2. a2<\/sup> + b2<\/sup> = x2<\/sup><\/td>\n2. _____________________________<\/td>\n<\/tr>\n
3. c2<\/sup> < r2<\/sup><\/td>\n3. _____________________________<\/td>\n<\/tr>\n
4. c < x<\/td>\n4. Take the positive square root of each side.<\/td>\n<\/tr>\n
5. m \u2220 R = 90\u00b0<\/td>\n5. _____________________________<\/td>\n<\/tr>\n
6. m \u2220 C < m \u2220 R<\/td>\n6. Converse of the Hinge Theorem (Theorem 6.13)<\/td>\n<\/tr>\n
7. m \u2220 C < 90\u00b0<\/td>\n7. _____________________________<\/td>\n<\/tr>\n
8. \u2220C is an acute angle.<\/td>\n8. _____________________________<\/td>\n<\/tr>\n
9. \u2206ABC is an acute triangle.<\/td>\n9. _____________________________<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Answer:<\/p>\n

Question 43.
\nPROVING A THEOREM<\/strong>
\nProve the Pythagorean Inequalities Theorem (Theorem 9.3) when c2<\/sup> > a2<\/sup> + b2<\/sup>. (Hint: Look back at Exercise 42.)
\nAnswer:
\n\"Big
\n\"Big<\/p>\n

Maintaining Mathematical Proficiency<\/p>\n

Simplify the expression by rationalizing the denominator.<\/p>\n

Question 44.
\n\\(\\frac{7}{\\sqrt{2}}\\)<\/p>\n

Answer:
\n7\/\u221a2 = 7\u221a2 \/2.<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\n\\(\\frac{7}{\\sqrt{2}}\\) = 7\/\u221a2.
\n7\/\u221a2 = 7\/\u221a2\u00a0 x \u221a2 \/\u221a2 .
\n7 \u221a2 \/\u221a4.
\n7\u221a2 \/2.<\/p>\n

Question 45.
\n\\(\\frac{14}{\\sqrt{3}}\\)
\nAnswer:
\n\"Big<\/p>\n

Question 46.
\n\\(\\frac{8}{\\sqrt{2}}\\)<\/p>\n

Answer:
\n8\/\u221a2 = 8\u221a2 \/2.<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\n\\(\\frac{8}{\\sqrt{2}}\\) = 8\/\u221a2.
\n8\/\u221a2 = 8\/\u221a2\u00a0 x \u221a2 \/\u221a2 .
\n8 \u221a2 \/\u221a4.
\n8\u221a2 \/2.<\/p>\n

Question 47.
\n\\(\\frac{12}{\\sqrt{3}}\\)
\nAnswer:
\n\"Big<\/p>\n

9.2 Special Right Triangles<\/h3>\n

Exploration 1<\/strong><\/p>\n

Side Ratios of an Isosceles Right Triangle<\/p>\n

Work with a partner:<\/p>\n

\"Big<\/p>\n

a. Use dynamic geometry software to construct an isosceles right triangle with a leg length of 4 units.
\nAnswer:<\/p>\n

b. Find the acute angle measures. Explain why this triangle is called a 45\u00b0 – 45\u00b0 – 90\u00b0 triangle.
\nAnswer:<\/p>\n

c. Find the exact ratios of the side lenghts (using square roots).
\n\\(\\frac{A B}{A C}\\) = ____________
\n\\(\\frac{A B}{B C}\\) = ____________
\n\\(\\frac{A B}{B C}\\) = ____________
\nATTENDING TO PRECISION<\/strong>
\nTo be proficient in math, you need to express numerical answers with a degree of precision appropriate for the problem context.
\nAnswer:<\/p>\n

d. Repeat parts (a) and (c) for several other isosceles right triangles. Use your results to write a conjecture about the ratios of the side lengths of an isosceles right triangle.
\nAnswer:<\/p>\n

Exploration 2<\/strong><\/p>\n

Work with a partner.<\/p>\n

a. Use dynamic geometry software to construct a right triangle with acute angle measures of 30\u00b0 and 60\u00b0 (a 30\u00b0 – 60\u00b0 – 90\u00b0 triangle), where the shorter leg length is 3 units.<\/p>\n

b. Find the exact ratios of the side lengths (using square roots).
\n\\(\\frac{A B}{A C}\\) = ____________
\n\\(\\frac{A B}{B C}\\) = ____________
\n\\(\\frac{A B}{B C}\\) = ____________
\n\"Big
\nAnswer:<\/p>\n

C. Repeat parts (a) and (b) for several other 30\u00b0 – 60\u00b0 – 90\u00b0 triangles. Use your results to write a conjecture about the ratios of the side lengths of a 30\u00b0 – 60\u00b0 – 90\u00b0 triangle.
\nAnswer:<\/p>\n

Communicate Your Answer<\/p>\n

Question 3.
\nWhat is the relationship among the side lengths of 45\u00b0- 45\u00b0 – 90\u00b0 triangles? 30\u00b0 – 60\u00b0 – 90\u00b0 triangles?
\nAnswer:<\/p>\n

Lesson 9.2 Special Right Triangles<\/h3>\n

Monitoring Progress<\/strong><\/p>\n

Find the value of the variable. Write your answer in simplest form.<\/p>\n

Question 1.
\n\"Big<\/p>\n

Answer:
\nx = 4<\/p>\n

Explanation:
\n(2\u221a2)\u00b2 = x\u00b2 + x\u00b2
\n8 = 2x\u00b2
\nx\u00b2 = 4
\nx = 4<\/p>\n

Question 2.
\n\"Big<\/p>\n

Answer:
\ny = 2<\/p>\n

Explanation:
\ny\u00b2 = 2 + 2
\ny\u00b2 = 4
\ny = 2<\/p>\n

Question 3.
\n\"Big<\/p>\n

Answer:
\nx = 3, y = 2\u221a3<\/p>\n

Explanation:
\nlonger leg = shorter leg \u2022 \u221a3
\nx = \u221a3 \u2022 \u221a3
\nx = 3
\nhypotenuse = shorter leg \u2022 2
\n= \u221a3 \u2022 2 = 2\u221a3<\/p>\n

Question 4
\n\"Big<\/p>\n

Answer:
\nh = 2\u221a3<\/p>\n

Explanation:
\nlonger leg = shorter leg \u2022 \u221a3
\nh = 2\u221a3<\/p>\n

Question 5.
\nThe logo on a recycling bin resembles an equilateral triangle with side lengths of 6 centimeters. Approximate the area of the logo.<\/p>\n

Answer:
\nArea is \\(\\frac { 1 }{ 3\u221a3 } \\)<\/p>\n

Explanation:
\nArea = \\(\\frac { \u221a3 }{ 4 } \\) a\u00b2
\n= \\(\\frac { \u221a3 }{ 4 } \\)(6)\u00b2
\n= \\(\\frac { 1 }{ 3\u221a3 } \\)<\/p>\n

Question 6.
\nThe body of a dump truck is raised to empty a load of sand. How high is the 14-foot-long body from the frame when it is tipped upward by a 60\u00b0 angle?
\n\"Big<\/p>\n

Answer:
\n28\/3 ft high is the 14-foot-long body from the frame when it is tipped upward by a 60\u00b0 angle.<\/p>\n

Explanation:
\nHeight of body at 90 degrees = 14 ft
\nHeight of body at 1 degree = 14\/90
\nHeight of body at 60 degrees = 14 x 60\/90
\n= 14 x 2\/3
\n= 28\/3 ft<\/p>\n

Exercise 9.2 Special Right Triangles<\/h3>\n

Vocabulary and Core Concept Check<\/p>\n

Question 1.
\nVOCABULARY<\/strong>
\nName two special right triangles by their angle measures.
\nAnswer:
\n\"Big<\/p>\n

Question 2.
\nWRITING<\/strong>
\nExplain why the acute angles in an isosceles right triangle always measure 45\u00b0.<\/p>\n

Answer:
\nBecause the acute angles of a right isosceles triangle must be congruent by the base angles theorem and complementary, their measures must be 90\u00b0\/2 = 45\u00b0.<\/p>\n

Monitoring Progress and Modeling with Mathematics<\/p>\n

In Exercises 3 – 6, find the value of x. Write your answer in simplest form.<\/p>\n

Question 3.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 4.
\n\"Big<\/p>\n

Answer:
\nx = 10<\/p>\n

Explanation:
\nhypotenuse = leg \u2022 \u221a2
\nx = 5\u221a2 \u2022 \u221a2
\nx = 10<\/p>\n

Question 5.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 6.
\n\"Big<\/p>\n

Answer:
\nx = \\(\\frac { 9 }{ \u221a2 } \\)<\/p>\n

Explanation:
\nhypotenuse = leg \u2022 \u221a2
\n9 = x \u2022 \u221a2
\nx = \\(\\frac { 9 }{ \u221a2 } \\)<\/p>\n

In Exercises 7 – 10, find the values of x and y. Write your answers in simplest form.<\/p>\n

Question 7.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 8.
\n\"Big<\/p>\n

Answer:
\nx = 3, y = 6<\/p>\n

Explanation:
\nhypotenuse = 2 \u2022 shorter leg
\ny = 2 \u2022 3
\ny = 6
\nlonger leg = \u221a3 \u2022 shorter leg
\n3\u221a3 = \u221a3x
\nx = 3<\/p>\n

Question 9.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 10.
\n\"Big<\/p>\n

Answer:
\nx = 18, y = 6\u221a3<\/p>\n

Explanation:
\nhypotenuse = 2 \u2022 shorter leg
\n12\u221a3 = 2y
\ny = 6\u221a3
\nlonger leg = \u221a3 \u2022 shorter leg
\nx = \u221a3 . 6\u221a3
\nx = 18<\/p>\n

ERROR ANALYSIS<\/strong>
\nIn Exercises 11 and 12, describe and correct the error in finding the length of the hypotenuse.<\/p>\n

Question 11.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 12.
\n\"Big<\/p>\n

Answer:
\nhypotenuse = leg \u2022 \u221a2 = \u221a5 . \u221a2 = \u221a10<\/p>\n

In Exercises 13 and 14. sketch the figure that is described. Find the indicated length. Round decimal answers to the nearest tenth.<\/p>\n

Question 13.
\nThe side length of an equilateral triangle is 5 centimeters. Find the length of an altitude.
\nAnswer:
\n\"Big<\/p>\n

Question 14.
\nThe perimeter of a square is 36 inches. Find the length of a diagonal.<\/p>\n

Answer:
\nThe length of a diagonal is 9\u221a2<\/p>\n

Explanation:
\nSide of the square = 36\/4 = 9
\nsquare diagonal = \u221a2a = \u221a2(9) = 9\u221a2<\/p>\n

In Exercises 15 and 16, find the area of the figure. Round decimal answers to the nearest tenth.<\/p>\n

Question 15.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 16.
\n\"Big<\/p>\n

Answer:
\nArea is 40\u221a(1\/3) sq m<\/p>\n

Explanation:
\nlonger leg = \u221a3 \u2022 shorter leg
\n4 = \u221a3 \u2022 shorter leg
\nshorter leg = 4\/\u221a3
\nh\u00b2 = 16\/3 + 16
\nh\u00b2 = 16(4\/3)
\nh = 8\u221a(1\/3)
\nArea of the parallelogram = 5(8\u221a(1\/3)) = 40\u221a(1\/3) sq m<\/p>\n

Question 17.
\nPROBLEM SOLVING<\/strong>
\nEach half of the drawbridge is about 284 feet long. How high does the drawbridge rise when x is 30\u00b0? 45\u00b0? 60\u00b0?
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 18.
\nMODELING WITH MATHEMATICS<\/strong>
\nA nut is shaped like a regular hexagon with side lengths of 1 centimeter. Find the value of x. (Hint: A regular hexagon can be divided into six congruent triangles.)
\n\"Big<\/p>\n

Answer:
\nSide length = 1cm
\nA regular hexagon has six equal the side length. A line drawn from the centre to any vertex will have the same length as any side.
\nThis implies the radius is equal to the side length.
\nAs a result, when lines are drawn from the centre to each of the vertexes, a
\nregular hexagon is said to be made of six equilateral triangles.
\nFrom the diagram, x = 2\u00d7 apothem
\nApothem is the distance from the centre of a regular polygon to the midpoint of side.
\nUsing Pythagoras theorem, we would get the apothem
\nHypotenuse\u00b2 = opposite\u00b2 + adjacent\u00b2
\n1\u00b2 = apothem\u00b2 + (\u00bd)\u00b2
\nApothem = \u221a(1\u00b2 -(\u00bd)\u00b2)
\n= \u221a(1-\u00bc) = \u221a\u00be
\nApothem = \u00bd\u221a3
\nx = 2\u00d7 Apothem = 2 \u00d7 \u00bd\u221a3
\nx = \u221a3<\/p>\n

Question 19.
\nPROVING A THEOREM<\/strong>
\nWrite a paragraph proof of the 45\u00b0- 45\u00b0- 90\u00b0 Triangle Theorem (Theorem 9.4).
\nGiven \u2206DEF is a 45\u00b0 – 45\u00b0 – 90\u00b0 triangle.
\nProve The hypotenuse is \u221a2 times as long as each leg.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 20.
\nHOW DO YOU SEE IT?<\/strong>
\nThe diagram shows part of the wheel of Theodorus.
\n\"Big
\na. Which triangles, if any, are 45\u00b0 – 45\u00b0 – 90\u00b0 triangles?<\/p>\n

Answer:<\/p>\n

b. Which triangles, if any, are 30\u00b0 – 60\u00b0 – 90\u00b0 triangles?<\/p>\n

Answer:<\/p>\n

Question 21.
\nPROVING A THEOREM<\/strong>
\nWrite a paragraph proof of the 30\u00b0 – 60\u00b0 – 90\u00b0 Triangle Theorem (Theorem 9.5).
\n(Hint: Construct \u2206JML congruent to \u2206JKL.)
\nGiven \u2206JKL is a 30\u00b0 60\u00b0 9o\u00b0 triangle.
\nProve The hypotenuse is twice as long as the shorter leg, and the longer leg is \u221a3 times as long as the shorter leg.
\n\"Big
\nAnswer:
\n\"Big
\n\"Big<\/p>\n

Question 22.
\nTHOUGHT PROVOKING<\/strong>
\nA special right triangle is a right triangle that has rational angle measures and each side length contains at most one square root. There are only three special right triangles. The diagram below is called the Ailles rectangle. Label the sides and angles in the diagram. Describe all three special right triangles.
\n\"Big<\/p>\n

Answer:
\n\"Big<\/p>\n

Question 23.
\nWRITING<\/strong>
\nDescribe two ways to show that all isosceles right triangles are similar to each other.
\nAnswer:
\n\"Big<\/p>\n

Question 24.
\nMAKING AN ARGUMENT<\/strong>
\nEach triangle in the diagram is a 45\u00b0 – 45\u00b0 – 90\u00b0 triangle. At Stage 0, the legs of the triangle are each 1 unit long. Your brother claims the lengths of the legs of the triangles added are halved at each stage. So, the length of a leg of a triangle added in Stage 8 will be \\(\\frac{1}{256}\\) unit. Is your brother correct? Explain your reasoning.
\n\"Big
\nAnswer:<\/p>\n

Question 25.
\nUSING STRUCTURE<\/strong>
\n\u0394TUV is a 30\u00b0 – 60\u00b0 – 90\u00b0 triangle. where two vertices are U(3, – 1) and V( – 3, – 1), \\(\\overline{U V}\\) is the hypotenuse. and point T is in Quadrant I. Find the coordinates of T.
\nAnswer:
\n\"Big
\n\"Big<\/p>\n

Maintaining Mathematical Proficiency<\/p>\n

Find the Value of x.<\/p>\n

Question 26.
\n\u0394DEF ~ \u0394LMN
\n\"Big<\/p>\n

Answer:
\nx = 18<\/p>\n

Explanation:
\n\\(\\frac { DE }{ LM } \\) = \\(\\frac { DF }{ LN } \\)
\n\\(\\frac { 12 }{ x } \\) = \\(\\frac { 20 }{ 30 } \\)
\n\\(\\frac { 12 }{ x } \\) = \\(\\frac { 2 }{ 3 } \\)
\nx = 12(\\(\\frac { 3 }{ 2 } \\)) = 18<\/p>\n

Question 27.
\n\u0394ABC ~ \u0394QRS
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

9.3 Similar Right Triangles<\/h3>\n

Exploration 1<\/strong><\/p>\n

Writing a Conjecture<\/p>\n

a. Use dynamic geometry software to construct right \u2206ABC, as shown. Draw \\(\\overline{C D}\\) so that it is an altitude from the right angle to the hypotenuse of \u2206ABC.
\n\"Big
\nAnswer:<\/p>\n

b. The geometric mean of two positive numbers a and b is the positive number x that satisfies
\n\\(\\frac{a}{x}=\\frac{x}{b}\\)
\nx is the geometric mean of a and b.
\nWrite a proportion involving the side lengths of \u2206CBD and \u2206ACD so that CD is the geometric mean of two of the other side lengths. Use similar triangles to justify your steps.
\nAnswer:<\/p>\n

c. Use the proportion you wrote in part (b) to find CD.
\nAnswer:<\/p>\n

d. Generalize the proportion you wrote in part (b). Then write a conjecture about how the geometric mean is related to the altitude from the right angle to the hypotenuse of a right triangle.
\nCONSTRUCTING VIABLE ARGUMENTS<\/strong>
\nTo be proficient in math, you need to understand and use stated assumptions, definitions, and previously established results in constructing arguments.
\nAnswer:<\/p>\n

Exploration 2<\/strong><\/p>\n

Comparing Geometric and Arithmetic Means<\/p>\n

Work with a partner:
\nUse a spreadsheet to find the arithmetic mean and the geometric mean of several pairs of positive numbers. Compare the two means. What do you notice?
\n\"Big
\nAnswer:<\/p>\n

Communicate Your Answer<\/p>\n

Question 3.
\nHow are altitudes and geometric means of right triangles related?
\nAnswer:<\/p>\n

Lesson 9.3 Similar Right Triangles<\/h3>\n

Monitoring progress<\/strong><\/p>\n

Identify the similar triangles.<\/p>\n

Question 1.
\n\"Big<\/p>\n

Answer:
\n\u25b3QRS ~ \u25b3 QST<\/p>\n

Question 2.
\n\"Big
\nAnswer:
\n\u25b3EFG ~ \u25b3 EHG<\/p>\n

Question 3.
\n\"Big<\/p>\n

Answer:
\n\u25b3EGH ~ \u25b3EFG
\n\\(\\frac { EF }{ EG } \\) = \\(\\frac { GF }{ GH } \\)
\n\\(\\frac { 5 }{ 3 } \\) = \\(\\frac { 4 }{ x } \\)
\nx = 2.4<\/p>\n

Question 4.
\n\"Big
\nAnswer:
\n\u25b3JLM ~ \u25b3LMK
\n\\(\\frac { JL }{ LM } \\) = \\(\\frac { JM }{ KM } \\)
\n\\(\\frac { 13 }{ 5 } \\) = \\(\\frac { 12 }{ x } \\)
\nx = 4.615<\/p>\n

Find the geometric mean of the two numbers.<\/p>\n

Question 5.
\n12 and 27<\/p>\n

Answer:
\nx = \u221a(12 x 27)
\nx = \u221a324 = 18<\/p>\n

Question 6.
\n18 and 54<\/p>\n

Answer:
\nx = \u221a(18 x 54) = \u221a(972)
\nx = 31.17<\/p>\n

Question 7.
\n16 and 18<\/p>\n

Answer:
\nx = \u221a(16 x 18) = \u221a(288)
\nx = 16.970<\/p>\n

Question 8.
\nFind the value of x in the triangle at the left.
\n\"Big<\/p>\n

Answer:
\nx = \u221a(9 x 4)
\nx = 6<\/p>\n

Question 9.
\nWHAT IF?<\/strong>
\nIn Example 5, the vertical distance from the ground to your eye is 5.5 feet and the distance from you to the gym wall is 9 feet. Approximate the height of the gym wall.<\/p>\n

Answer:
\n9\u00b2 = 5.5 x w
\n81 = 5.5 x w
\nw = 14.72
\nThe height of the wall = 14.72 + 5.5 = 20.22<\/p>\n

Exercise 9.3 Similar Right Triangles<\/h3>\n

Vocabulary and Core Concept Check<\/p>\n

Question 1.
\nCOMPLETE THE SENTENCE<\/strong>
\nIf the altitude is drawn to the hypotenuse of a right triangle, then the two triangles formed are similar to the original triangle and ____________ .
\nAnswer:
\n\"Big<\/p>\n

Question 2.
\nWRITING<\/strong>
\nIn your own words, explain geometric mean.<\/p>\n

Answer:
\nThe geometric mean is the average value or mean that signifies the central tendency of set of numbers by finding the product of their values.<\/p>\n

Monitoring progress and Modeling with Mathematics<\/p>\n

In Exercises 3 and 4, identify the similar triangles.<\/p>\n

Question 3.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 4.
\n\"Big<\/p>\n

Answer:
\n\u25b3LKM ~ \u25b3LMN ~ \u25b3MKN<\/p>\n

In Exercises 5 – 10, find the value of x.<\/p>\n

Question 5.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 6.
\n\"Big<\/p>\n

Answer:
\nx = 9.6<\/p>\n

Explanation:
\n\\(\\frac { QR }{ SR } \\) = \\(\\frac { SR }{ TS } \\)
\n\\(\\frac { 20 }{ 16 } \\) = \\(\\frac { 12 }{ x } \\)
\n1.25 = \\(\\frac { 12 }{ x } \\)
\nx = 9.6<\/p>\n

Question 7.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 8.
\n\"Big<\/p>\n

Answer:
\nx = 14.11<\/p>\n

Explanation:
\n\\(\\frac { AB }{ AC } \\) = \\(\\frac { BD }{ BC } \\)
\n\\(\\frac { 16 }{ 34 } \\) = \\(\\frac { x }{ 30 } \\)
\nx = 14.11<\/p>\n

Question 9.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 10.
\n\"Big<\/p>\n

Answer:
\nx = 2.77<\/p>\n

Explanation:
\n\\(\\frac { 5.8 }{ 3.5 } \\) = \\(\\frac { 4.6 }{ x } \\)
\nx = 2.77<\/p>\n

In Exercises 11 – 18, find the geometric mean of the two numbers.<\/p>\n

Question 11.
\n8 and 32
\nAnswer:
\n\"Big<\/p>\n

Question 12.
\n9 and 16<\/p>\n

Answer:
\nx = \u221a(9 x 16)
\nx = 12<\/p>\n

Question 13.
\n14 and 20
\nAnswer:
\n\"Big<\/p>\n

Question 14.
\n25 and 35<\/p>\n

Answer:
\nx = \u221a(25 x 35)
\nx = 29.5<\/p>\n

Question 15.
\n16 and 25
\nAnswer:
\n\"Big<\/p>\n

Question 16.
\n8 and 28<\/p>\n

Answer:
\nx = \u221a(8 x 28)
\nx = 14.96<\/p>\n

Question 17.
\n17 and 36
\nAnswer:
\n\"Big<\/p>\n

Question 18.
\n24 and 45<\/p>\n

Answer:
\nx = \u221a(24 x 45)
\nx = 32.86<\/p>\n

In Exercises 19 – 26. find the value of the variable.<\/p>\n

Question 19.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 20.
\n\"Big<\/p>\n

Answer:
\ny = \u221a(5 x 8)
\ny = \u221a40
\ny = 2\u221a10<\/p>\n

Question 21.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 22.
\n\"Big<\/p>\n

Answer:
\n10 \u2022 10 = 25 \u2022 x
\n100 = 25x
\nx = 4<\/p>\n

Question 23.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 24.
\n\"Big<\/p>\n

Answer:
\nb\u00b2 = 16(16 + 6)
\nb\u00b2 = 16(22) = 352
\nb = 18.76<\/p>\n

Question 25.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 26.
\n\"Big<\/p>\n

Answer:
\nx\u00b2 = 8(8 + 2)
\nx\u00b2 = 8(10) = 80
\nx = 8.9<\/p>\n

ERROR ANALYSIS<\/strong>
\nIn Exercises 27 and 28, describe and correct the error in writing an equation for the given diagram.<\/p>\n

Question 27.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 28.
\n\"Big<\/p>\n

Answer:
\nd\u00b2 = g \u2022 e<\/p>\n

MODELING WITH MATHEMATICS<\/strong>
\nIn Exercises 29 and 30, use the diagram.<\/p>\n

\"Big<\/p>\n

Question 29.
\nYou want to determine the height of a monument at a local park. You use a cardboard square to line up the top and bottom of the monument, as shown at the above left. Your friend measures the vertical distance from the ground to your eye and the horizontal distance from you to the monument. Approximate the height of the monument.
\nAnswer:
\n\"Big<\/p>\n

Question 30.
\nYour classmate is standing on the other side of the monument. She has a piece of rope staked at the base of the monument. She extends the rope to the cardboard square she is holding lined up to the top and bottom of the monument. Use the information in the diagram above to approximate the height of the monument. Do you get the same answer as in Exercise 29? Explain your reasoning.<\/p>\n

Answer:<\/p>\n

MATHEMATICAL CONNECTIONS<\/strong>
\nIn Exercises 31 – 34. find the value(s) of the variable(s).<\/p>\n

Question 31.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 32.
\n\"Big<\/p>\n

Answer:
\n\\(\\frac { 6 }{ b + 3 } \\) = \\(\\frac { 8 }{ 6 } \\)
\n36 = 8(b + 3)
\n36 = 8b + 24
\n8b = 12
\nb = \\(\\frac { 3 }{ 2 } \\)<\/p>\n

Question 33.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 34.
\n\"Big<\/p>\n

Answer:
\nx = 42.66, y = 40, z = 53<\/p>\n

Explanation:
\n\\(\\frac { 24 }{ 32 } \\) = \\(\\frac { 32 }{ x } \\)
\n0.75 = \\(\\frac { 32 }{ x } \\)
\nx = 42.66
\ny\u00a0= \u221a24\u00b2 + 32\u00b2
\ny\u00a0= \u221a576 + 1024 = 40
\nz = \u221a42.66\u00b2 + 32\u00b2 = \u221a1819.87 + 1024 = 53<\/p>\n

Question 35.
\nREASONING<\/strong>
\nUse the diagram. Decide which proportions are true. Select all that apply.
\n\"Big
\n(A) \\(\\frac{D B}{D C}=\\frac{D A}{D B}\\)
\n(B) \\(\\frac{B A}{C B}=\\frac{C B}{B D}\\)
\n(C) \\(\\frac{C A}{B \\Lambda}=\\frac{B A}{C A}\\)
\n(D) \\(\\frac{D B}{B C}=\\frac{D A}{B A}\\)
\nAnswer:
\n\"Big<\/p>\n

Question 36.
\nANALYZING RELATIONSHIPS<\/strong>
\nYou are designing a diamond-shaped kite. You know that AD = 44.8 centimeters, DC = 72 centimeters, and AC = 84.8 centimeters. You Want to use a straight crossbar \\(\\overline{B D}\\). About how long should it be? Explain your reasoning.
\n\"Big<\/p>\n

Answer:
\nBD = 76.12<\/p>\n

Explanation:
\nAD = 44.8 cm, DC = 72 cm, and AC = 84.8 cm
\nTwo disjoint pairs of consecutive sides are congruent.
\nSo, AD = AB = 44.8 cm
\nDC = BC = 72 cm
\nThe diagonals are perpendicular.
\nSo, AC \u22a5 BD
\nAC = AO + OC
\nAX = x + y = 84.8 — (i)
\nPerpendicular bisects the diagonal BD into equal parts let it be z.
\nBD = BO + OD
\nBD = z + z
\nUsing pythagorean theorem
\n44.8\u00b2 = x\u00b2 + z\u00b2 —- (ii)
\n72\u00b2 = y\u00b2 + z\u00b2 —– (iii)
\nSubtract (ii) and (iii)
\n72\u00b2 – 44.8\u00b2 = y\u00b2+ z\u00b2 – x\u00b2 – z\u00b2
\n5184 – 2007.04 = (x + y) (x – y)
\n3176.96 = (84.8)(x – y)
\n37.464 = x – y —- (iv)
\nAdd (i) & (iv)
\nx + y + x – y = 84.8 + 37.464
\n2x = 122.264
\nx = 61.132
\nx + y = 84.8
\n61.132 + y = 84.8
\ny = 23.668
\n44.8\u00b2 = x\u00b2 + z\u00b2
\nz = 38.06
\nBD = z + z
\nBD = 76.12<\/p>\n

Question 37.
\nANALYZING RELATIONSHIPS<\/strong>
\nUse the Geometric Mean Theorems (Theorems 9.7 and 9.8) to find AC and BD.
\n\"Big
\nAnswer:
\n\"Big
\n\"Big<\/p>\n

Question 38.
\nHOW DO YOU SEE IT?<\/strong>
\nIn which of the following triangles does the Geometric Mean (Altitude) Theorem (Theorem 9.7) apply?
\n(A)
\n\"Big
\n(B)
\n\"Big
\n(C)
\n\"Big
\n(D)
\n\"Big<\/p>\n

Answer:<\/p>\n

Question 39.
\nPROVING A THEOREM<\/strong>
\nUse the diagram of \u2206ABC. Copy and complete the proof of the Pythagorean Theorem (Theorem 9. 1).
\n\"Big
\nGiven In \u2206ABC, \u2206BCA is a right angle.
\nProve c2<\/sup> = a2<\/sup> + b2<\/sup><\/p>\n\n\n\n\n\n\n\n\n\n\n\n\n
Statements<\/td>\nReasons<\/td>\n<\/tr>\n
1. In \u2206ABC, \u2220BCA is a right angle.<\/td>\n1. ________________________________<\/td>\n<\/tr>\n
2. Draw a perpendicular segment (altitude) from C to \\(\\overline{A B}\\).<\/td>\n2. Perpendicular Postulate (Postulate 3.2)<\/td>\n<\/tr>\n
3. ce = a2<\/sup> and cf = b2<\/sup><\/td>\n3. ________________________________<\/td>\n<\/tr>\n
4. ce + b2<\/sup> = ___\u00a0 + b2<\/sup><\/td>\n4. Addition Property of Equality<\/td>\n<\/tr>\n
5. ce + cf = a2<\/sup> + b2<\/sup><\/td>\n5. ________________________________<\/td>\n<\/tr>\n
6. c(e + f) a2<\/sup> + b2<\/sup><\/td>\n6. ________________________________<\/td>\n<\/tr>\n
7. e + f = ________<\/td>\n7. Segment Addition Postulate (Postulate 1.2)<\/td>\n<\/tr>\n
8. c\u00a0 \u2022 c = a2<\/sup> + b2<\/sup><\/td>\n8. ________________________________<\/td>\n<\/tr>\n
9. c2<\/sup> = a2<\/sup> + b2<\/sup><\/td>\n9. Simplify.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Answer:
\n\"Big
\n\"Big<\/p>\n

Question 40.
\nMAKING AN ARGUMENT<\/strong>
\nYour friend claims the geometric mean of 4 and 9 is 6. and then labels the triangle, as shown. Is your friend correct? Explain your reasoning.
\n\"Big<\/p>\n

Answer:
\nG.M = \u221a(4 x 9) = \u221a36 = 6
\nMy friend is correct.<\/p>\n

In Exercises 41 and 42, use the given statements to prove the theorem.<\/p>\n

Gien \u2206ABC is a right triangle.
\nAltitude \\(\\overline{C D}\\) is dravn to hypotenuse \\(\\overline{A B}\\).<\/p>\n

Question 41.
\nPROVING A THEOREM<\/strong>
\nProve the Geometric Mean (Altitude) Theorem (Theorem 9.7) b showing that CD2<\/sup> = AD \u2022 BD.
\nAnswer:
\n\"Big
\n\"Big<\/p>\n

Question 42.
\nPROVING A THEOREM<\/strong>
\nProve the Geometric Mean ( Leg) Theorem (Theorem 9.8) b showing that CB2<\/sup> = DB \u2022 AB and AC2<\/sup> = AD \u2022 AB.<\/p>\n

Answer:<\/p>\n

Question 43.
\nCRITICAL THINKING<\/strong>
\nDraw a right isosceles triangle and label the two leg lengths x. Then draw the altitude to the hypotenuse and label its length y. Now, use the Right Triangle Similarity Theorem (Theorem 9.6) to draw the three similar triangles from the image and label an side length that is equal to either x or y. What can you conclude about the relationship between the two smaller triangles? Explain your reasoning.
\nAnswer:
\n\"Big<\/p>\n

Question 44.
\nTHOUGHT PROVOKING<\/strong>
\nThe arithmetic mean and geometric mean of two nonnegative numbers x and y are shown.
\narithmetic mean = \\(\\frac{x+y}{2}\\)
\ngeometric mean = \\(\\sqrt{x y}\\)
\nWrite an inequality that relates these two means. Justify your answer.<\/p>\n

Answer:<\/p>\n

Question 45.
\nPROVING A THEOREM<\/strong>
\nProve the Right Triangle Similarity Theorem (Theorem 9.6) by proving three similarity statements.
\nGiven \u2206ABC is a right triangle.
\nAltitude \\(\\overline{C D}\\) is drawn to hvpotenuse \\(\\overline{A B}\\).
\nProve \u2206CBD ~ \u2206ABC, \u2206ACD ~ \u2206ABC,
\n\u2206CBD ~ \u2206ACD
\nAnswer:
\n\"Big
\n\"Big
\n\"Big<\/p>\n

Maintaining Mathematical proficiency<\/p>\n

Solve the equation for x.<\/p>\n

Question 46.
\n13 = \\(\\frac{x}{5}\\)<\/p>\n

Answer:
\n13 = \\(\\frac{x}{5}\\)
\nx = 65<\/p>\n

Question 47.
\n29 = \\(\\frac{x}{4}\\)
\nAnswer:
\n\"Big<\/p>\n

Question 48.
\n9 = \\(\\frac{78}{x}\\)<\/p>\n

Answer:
\n9 = \\(\\frac{78}{x}\\)
\n9x = 78
\nx = 8.6<\/p>\n

Question 49.
\n30 = \\(\\frac{115}{x}\\)
\nAnswer:
\n\"Big<\/p>\n

9.1 to 9.3 Quiz<\/h3>\n

Find the value of x. Tell whether the side lengths form a Pythagorean triple.<\/p>\n

Question 1.
\n\"Big
\nAnswer:
\nx = 15<\/p>\n

Explanation:
\nx\u00b2 = 9\u00b2 + 12\u00b2
\nx\u00b2 = 81 + 144
\nx\u00b2 = 225
\nx = 15<\/p>\n

Question 2.
\n\"Big<\/p>\n

Answer:
\nx = 10.63<\/p>\n

Explanation:
\nx\u00b2 = 7\u00b2 + 8\u00b2 = 49 + 64
\nx = \u221a113
\nx = 10.63<\/p>\n

Question 3.
\n\"Big
\nAnswer:
\nx = 4\u221a3<\/p>\n

Explanation:
\n8\u00b2 = x\u00b2 + 4\u00b2
\n64 = x\u00b2 + 16
\nx\u00b2 = 48
\nx = 4\u221a3<\/p>\n

Verify that the segment lengths form a triangle. Is the triangle acute, right, or obtuse?
\n(Section 9.1)
\nQuestion 4.
\n24, 32, and 40<\/p>\n

Answer:
\nTriangle is a right angle trinagle.<\/p>\n

Explanation:
\n40\u00b2 = 1600
\n24\u00b2 + 32\u00b2 = 576 + 1024 = 1600
\n40\u00b2 = 24\u00b2 + 32\u00b2
\nSo, the triangle is a right angle trinagle.<\/p>\n

Question 5.
\n7, 9, and 13<\/p>\n

Answer:
\nTriangle is an obtuse trinagle.<\/p>\n

Explanation:
\n13\u00b2 = 169
\n7\u00b2 + 9\u00b2 = 49 + 81 = 130
\n13\u00b2 > 7\u00b2 + 9\u00b2
\nSo, the triangle is an obtuse trinagle.<\/p>\n

Question 6.
\n12, 15, and 10\u221a3<\/p>\n

Answer:
\nTriangle is an acute trinagle.<\/p>\n

Explanation:
\n15\u00b2 = 225
\n12\u00b2 + (10\u221a3)\u00b2 = 144 + 300 = 444
\n15\u00b2 < 12\u00b2 + (10\u221a3)\u00b2
\nSo, the triangle is an acute trinagle.<\/p>\n

Find the values of x and y. Write your answers in the simplest form.<\/p>\n

Question 7.
\n\"Big<\/p>\n

Answer:
\nx = 6, y = 6\u221a2<\/p>\n

Explanation:
\nx = 6
\nhypotenuse = leg \u2022 \u221a2
\ny = 6\u221a2<\/p>\n

Question 8.
\n\"Big<\/p>\n

Answer:
\ny = 8\u221a3, x = 16<\/p>\n

Explanation:
\nlonger leg = shorter leg \u2022 \u221a3
\ny = 8\u221a3
\nx\u00b2 = 8\u00b2 + (8\u221a3)\u00b2 = 64 + 192
\nx = 16<\/p>\n

Question 9.
\n\"Big<\/p>\n

Answer:
\nx = 5\u221a2, y = 5\u221a6<\/p>\n

Explanation:
\nlonger leg = shorter leg \u2022 \u221a3
\ny = x\u221a3
\ny = 5\u221a6
\nhypotenuse = shorter leg \u2022 2
\n10\u221a2 = 2x
\nx = 5\u221a2<\/p>\n

Find the geometric mean of the two numbers.
\nQuestion 10.
\n6 and 12<\/p>\n

Answer:
\nG.M = \u221a(6 \u2022 12) = 6\u221a2<\/p>\n

Question 11.
\n15 and 20<\/p>\n

Answer:
\nG.M = \u221a(15 \u2022 20) = 10\u221a3<\/p>\n

Question 12.
\n18 and 26<\/p>\n

Answer:
\nG.M = \u221a(18 \u2022 26) = 6\u221a13<\/p>\n

Identify the similar right triangles. Then find the value of the variable.<\/p>\n

Question 13.
\n\"Big<\/p>\n

Answer:
\nx = \u221a(8 \u2022 4)
\nx = 4\u221a2<\/p>\n

Question 14.
\n\"Big<\/p>\n

Answer:
\ny = \u221a(9 \u2022 6) = 3\u221a6<\/p>\n

Question 15.
\n\"Big<\/p>\n

Answer:<\/p>\n

Question 16.
\nTelevision sizes are measured by the length of their diagonal. You want to purchase a television that is at least 40 inches. Should you purchase the television shown? Explain your reasoning.
\n\"Big<\/p>\n

Answer:
\nx\u00b2 = 20.25\u00b2 + 36\u00b2
\nx\u00b2 = 410.0625 + 1296 = 1706.0625
\nx = 41.30
\nYes, i will purchase the television.<\/p>\n

Question 17.
\nEach triangle shown below is a right triangle.
\n\"Big
\na. Are any of the triangles special right triangles? Explain your reasoning.
\nAnswer:
\nA is a similar triangle.<\/p>\n

b. List all similar triangles. if any.
\nAnswer:
\nB, C and D, E are similar triangles.<\/p>\n

c. Find the lengths of the altitudes of triangles B and C.
\nAnswer:
\nB altitude = \u221a(9 + 27) = 6
\nC altitude = \u221a(36 + 72) = 6\u221a3<\/p>\n

9.4 The Tangent Ratio<\/h3>\n

Exploration 1<\/strong><\/p>\n

Calculating a Tangent Ratio<\/p>\n

Work with a partner<\/p>\n

a. Construct \u2206ABC, as shown. Construct segments perpendicular to \\(\\overline{A C}\\) to form right triangles that share vertex A and arc similar to \u2206ABC with vertices, as shown.
\n\"Big
\nAnswer:<\/p>\n

b. Calculate each given ratio to complete the table for the decimal value of tan A for each right triangle. What can you Conclude?
\n\"Big
\nAnswer:<\/p>\n

Exploration 2<\/strong><\/p>\n

Using a calculator<\/p>\n

Work with a partner: Use a calculator that has a tangent key to calculate the tangent of 36.87\u00b0. Do you get the same result as in Exploration 1? Explain.
\nATTENDING TO PRECISION<\/strong>
\nTo be proficient in math, you need to express numerical answers with a degree of precision appropriate for the problem context.
\nAnswer:<\/p>\n

Communicate Your Answer<\/p>\n

Question 3.
\nRepeat Exploration 1 for \u2206ABC with vertices A(0, 0), B(8, 5), and C(8, 0). Construct the seven perpendicular segments so that not all of them intersect \\(\\overline{A C}\\) at integer values of x. Discuss your results.
\nAnswer:<\/p>\n

Question 4.
\nHow is a right triangle used to find the tangent of an acute angle? Is there a unique right triangle that must be used?
\nAnswer:<\/p>\n

Lesson 9.4 The Tangent Ratio<\/h3>\n

Monitoring progress<\/strong><\/p>\n

Find tan J and tan K. Write each answer as a fraction and as a decimal rounded to four places.<\/p>\n

Question 1.
\n\"Big<\/p>\n

Answer:
\ntan K = \\(\\frac { opposite side }{ adjacent side } \\) = \\(\\frac { JL }{ KL } \\)
\n= \\(\\frac { 32 }{ 24 } \\) = \\(\\frac { 4 }{ 3 } \\) = 1.33
\ntan J = \\(\\frac { KL }{ JL } \\) = \\(\\frac { 24 }{ 32 } \\) = \\(\\frac { 3 }{ 4 } \\) = 0.75<\/p>\n

Question 2.
\n\"Big<\/p>\n

Answer:
\ntan K = \\(\\frac { LJ }{ LK } \\) = \\(\\frac { 15 }{ 8 } \\)
\ntan J = \\(\\frac { LK }{ LJ } \\) = \\(\\frac { 8 }{ 15 } \\)<\/p>\n

Find the value of x. Round your answer to the nearest tenth.<\/p>\n

Question 3.
\n\"Big<\/p>\n

Answer:
\nTan 61 = \\(\\frac { 22 }{ x } \\)
\n1.804 = \\(\\frac { 22 }{ x } \\)
\nx = 12.1951<\/p>\n

Question 4.
\n\"Big<\/p>\n

Answer:
\ntan 56 = \\(\\frac { x }{ 13 } \\)
\n1.482 = \\(\\frac { x }{ 13 } \\)
\nx = 19.266<\/p>\n

Question 5.
\nWHAT IF?<\/strong>
\nIn Example 3, the length of the shorter leg is 5 instead of 1. Show that the tangent of 60\u00b0 is still equal to \u221a3.<\/p>\n

Answer:
\nlonger leg = shorter leg \u2022 \u221a3
\n= 5\u221a3
\ntan 60 = \\(\\frac { 5\u221a3 }{ 5 } \\)
\n= \u221a3<\/p>\n

Question 6.
\nYou are measuring the height of a lamppost. You stand 40 inches from the base of the lamppost. You measure the angle ot elevation from the ground to the top of the lamppost to be 70\u00b0. Find the height h of the lamppost to the nearest inch.
\n\"Big<\/p>\n

Answer:
\ntan 70 = \\(\\frac { h }{ 40 } \\)
\n2.7474 = \\(\\frac { h }{ 40 } \\)
\nh = 109.896 in<\/p>\n

Exercise 9.4 The Tangent Ratio<\/h3>\n

Vocabulary and Core Concept Check<\/p>\n

Question 1.
\nCOMPLETE THE SENTENCE<\/strong>
\nThe tangent ratio compares the length of _________ to the length of ___________ .
\nAnswer:
\n\"Big<\/p>\n

Question 2.
\nWRITING<\/strong>
\nExplain how you know the tangent ratio is constant for a given angle measure.<\/p>\n

Answer:
\nWhen two triangles are similar, the corresponding sides are proportional which makes the ratio constant for a given acute angle measurement.<\/p>\n

Monitoring Progress and Modeling with Mathematics<\/p>\n

In Exercises 3 – 6, find the tangents of the acute angles in the right triangle. Write each answer as a fraction and as a decimal rounded to four decimal places.<\/p>\n

Question 3.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 4.
\n\"Big<\/p>\n

Answer:
\ntan F = \\(\\frac { DE }{ EF } \\) = \\(\\frac { 24 }{ 7 } \\)
\ntan D = \\(\\frac { EF }{ DE } \\) = \\(\\frac { 7 }{ 24 } \\)<\/p>\n

Question 5.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 6.
\n\"Big<\/p>\n

Answer:
\ntan K = \\(\\frac { JL }{ LK } \\) = \\(\\frac { 3 }{ 5 } \\)
\ntan J = \\(\\frac { LK }{ JL } \\) = \\(\\frac { 5 }{ 3 } \\)<\/p>\n

In Exercises 7 – 10, find the value of x. Round your answer to the nearest tenth.<\/p>\n

Question 7.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 8.
\n\"Big<\/p>\n

Answer:
\ntan 27 = \\(\\frac { x }{ 15 } \\)
\n0.509 = \\(\\frac { x }{ 15 } \\)
\nx = 7.635<\/p>\n

Question 9.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 10.
\n\"Big<\/p>\n

Answer:
\ntan 37 = \\(\\frac { 6 }{ x } \\)
\n0.753 = \\(\\frac { 6 }{ x } \\)
\nx = 7.968<\/p>\n

ERROR ANALYSIS<\/strong>
\nIn Exercises 11 and 12, describe the error in the statement of the tangent ratio. Correct the error if possible. Otherwise, write not possible.<\/p>\n

Question 11.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 12.
\n\"Big<\/p>\n

Answer:<\/p>\n

In Exercises 13 and 14, use a special right triangle to find the tangent of the given angle measure.<\/p>\n

Question 13.
\n45\u00b0
\nAnswer:
\n\"Big<\/p>\n

Question 14.
\n30\u00b0<\/p>\n

Answer:
\ntan 30\u00b0 = \\(\\frac { 1 }{ \u221a3 } \\)<\/p>\n

Question 15.
\nMODELING WITH MATHEMATICS<\/strong>
\nA surveyor is standing 118 Feet from the base of the Washington Monument. The surveyor measures the angle of elevation from the ground to the top of the monument to be 78\u00b0. Find the height h of the Washington Monument to the nearest foot.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 16.
\nMODELING WITH MATHEMATICS<\/strong>
\nScientists can measure the depths of craters on the moon h looking at photos of shadows. The length of the shadow cast by the edge of a crater is 500 meters. The angle of elevation of the rays of the Sun is 55\u00b0. Estimate the depth d of the crater.
\n\"Big<\/p>\n

Answer:
\ntan 55 = \\(\\frac { d }{ 500 } \\)
\n1.428 = \\(\\frac { d }{ 500 } \\)
\nd = 714 m
\nThe depth of the crater is 714 m<\/p>\n

Question 17.
\nUSING STRUCTURE<\/strong>
\nFind the tangent of the smaller acute angle in a right triangle with side lengths 5, 12, and 13.
\nAnswer:
\n\"Big<\/p>\n

Question 18.
\nUSING STRUCTURE<\/strong>
\nFind the tangent 0f the larger acute angle in a right triangle with side lengths 3, 4, and 5.<\/p>\n

Answer:
\ntan x = \\(\\frac { 4 }{ 3 } \\)<\/p>\n

Question 19.
\nREASONING<\/strong>
\nHow does the tangent of an acute angle in a right triangle change as the angle measure increases? Justify your answer.
\nAnswer:
\n\"Big<\/p>\n

Question 20.
\nCRITICAL THINKING<\/strong>
\nFor what angle measure(s) is the tangent of an acute angle in a right triangle equal to 1? greater than 1? less than 1? Justify your answer.<\/p>\n

Answer:
\nIn order for the tangent of an angle to equal 1, the opposite and adjacent sides of a right triangle must be the same. This means the right triangle is an isosceles right triangle so the angles are 45 – 45 – 90. The acute angle must be 1. In order for the tangent to be greater than 1, the opposite side must be greater than the adjacent side. This means the angle must be between 45 and 90 degrees. If the tangent is less than 1, this means the opposite side must be smaller than the adjacent side. The acute angle must be between 0 and 45.<\/p>\n

Question 21.
\nMAKING AN ARGUMENT<\/strong>
\nYour family room has a sliding-glass door. You want to buy an awning for the door that will be just long enough to keep the Sun out when it is at its highest point in the sky. The angle of elevation of the rays of the Sun at this points is 70\u00b0, and the height of the door is 8 feet. Your sister claims you can determine how far the overhang should extend by multiplying 8 by tan 70\u00b0. Is your sister correct? Explain.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 22.
\nHOW DO YOU SEE IT?<\/strong>
\nWrite expressions for the tangent of each acute angle in the right triangle. Explain how the tangent of one acute angle is related to the tangent of the other acute angle. What kind of angle pair is \u2220A and \u2220B?
\n\"Big<\/p>\n

Answer:
\ntan A = \\(\\frac { BC }{ AC } \\) = \\(\\frac { a }{ b } \\)
\ntan B = \\(\\frac { AC }{ BC } \\) = \\(\\frac { b }{ a } \\)<\/p>\n

Question 23.
\nREASONING<\/strong>
\nExplain why it is not possible to find the tangent of a right angle or an obtuse angle.
\nAnswer:
\n\"Big<\/p>\n

Question 24.
\nTHOUGHT PROVOKING<\/strong>
\nTo create the diagram below. you begin with an isosceles right triangle with legs 1 unit long. Then the hypotenuse of the first triangle becomes the leg of a second triangle, whose remaining leg is 1 unit long. Continue the diagram Until you have constructed an angle whose tangent is \\(\\frac{1}{\\sqrt{6}}\\). Approximate the measure of this angle.<\/p>\n

Answer:<\/p>\n

Question 25.
\nPROBLEM SOLVING<\/strong>
\nYour class is having a class picture taken on the lawn. The photographer is positioned 14 feet away from the center of the class. The photographer turns 50\u00b0 to look at either end of the class.
\n\"Big
\na. What is the distance between the ends of the class?
\nb. The photographer turns another 10\u00b0 either way to see the end of the camera range. If each student needs 2 feet of space. about how many more students can fit at the end of each row? Explain.
\nAnswer:
\n\"Big<\/p>\n

Question 26.
\nPROBLEM SOLVING<\/strong>
\nFind the perimeter of the figure. where AC = 26, AD = BF, and D is the midpoint of \\(\\overline{A C}\\).
\n\"Big<\/p>\n

Answer:<\/p>\n

Maintaining Mathematical proficiency<\/p>\n

Find the value of x.<\/p>\n

Question 27.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 28.
\n\"Big
\nAnswer:
\nlonger side = shorter side \u2022 \u221a3
\n7 = x\u221a3
\nx = \\(\\frac { 7 }{ \u221a3 } \\)
\nx = 4.04<\/p>\n

Question 29.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

9.5 The Sine and Cosine Ratios<\/h3>\n

Exploration 1<\/strong><\/p>\n

Work with a partner: Use dynamic geometry software.<\/p>\n

a. Construct \u2206ABC, as shown. Construct segments perpendicular to \\(\\overline{A C}\\) to form right triangles that share vertex A arid are similar to \u2206ABC with vertices, as shown.
\n\"Big
\nAnswer:<\/p>\n

b. Calculate each given ratio to complete the table for the decimal values of sin A and cos A for each right triangle. What can you conclude?
\n\"Big
\nAnswer:<\/p>\n

Communicate Your Answer<\/p>\n

Question 2.
\nHow is a right triangle used to find the sine and cosine of an acute angle? Is there a unique right triangle that must be used?
\nAnswer:<\/p>\n

Question 3.
\nIn Exploration 1, what is the relationship between \u2220A and \u2220B in terms of their measures\u2019? Find sin B and cos B. How are these two values related to sin A and cos A? Explain why these relationships exist.
\nLOOKING FOR STRUCTURE<\/strong>
\nTo be proficient in math, you need to look closely to discern a pattern or structure.
\nAnswer:<\/p>\n

Lesson 9.5 The Sine and Cosine Ratios<\/h3>\n

Monitoring Progress<\/strong><\/p>\n

Question 1.
\nFind sin D, sin F, cos D, and cos F. Write each answer as a fraction and as a decimal rounded to four places.
\n\"Big
\nAnswer:
\nsin D = \\(\\frac { 7 }{ 25 } \\)
\nsin F = \\(\\frac { 24 }{ 25 } \\)
\ncos D = \\(\\frac { 24 }{ 25 } \\)
\ncos F = \\(\\frac { 7 }{ 25 } \\)<\/p>\n

Explanation:
\nsin D = \\(\\frac { EF }{ DF } \\) = \\(\\frac { 7 }{ 25 } \\)
\nsin F = \\(\\frac { DE }{ DF } \\) = \\(\\frac { 24 }{ 25 } \\)
\ncos D = \\(\\frac { DE }{ DF } \\) = \\(\\frac { 24 }{ 25 } \\)
\ncos F = \\(\\frac { EF }{ DF } \\) = \\(\\frac { 7 }{ 25 } \\)<\/p>\n

Question 2.
\nWrite cos 23\u00b0 in terms of sine.<\/p>\n

Answer:
\ncos X = sin(90 – X)
\ncos 23\u00b0 = sin (90 – 23)
\n= sin(67)
\nSo, cos 23\u00b0 = sin 67\u00b0<\/p>\n

Question 3.
\nFind the values of u and t using sine and cosine. Round your answers to the nearest tenth.
\n\"Big
\nAnswer:
\nt = 7.2, u = 3.3<\/p>\n

Explanation:
\nsin 65 = \\(\\frac { t }{ 8 } \\)
\n0.906 = \\(\\frac { t }{ 8 } \\)
\nt = 7.2
\ncos 65 = \\(\\frac { u }{ 8 } \\)
\n0.422 x 8 = u
\nu = 3.3<\/p>\n

Question 4.
\nFind the sine and cosine of a 60\u00b0 angle.<\/p>\n

Answer:
\n\"Big
\nsin 60\u00b0 = \\(\\frac { \u221a3 }{ 2 } \\)
\ncos 60\u00b0 = \\(\\frac { 1 }{ 2 } \\)<\/p>\n

Question 5.
\nWHAT IF?<\/strong>
\nIn Example 6, the angle of depression is 28\u00b0. Find the distance x you ski down the mountain to the nearest foot.<\/p>\n

Answer:
\nsin 28\u00b0 = \\(\\frac { 1200 }{ x } \\)
\nx\u00a0= \\(\\frac { 1200 }{ 0.469 } \\)
\nx = 2558.6<\/p>\n

Exercise 9.5 The Sine and Cosine Ratios<\/h3>\n

Vocabulary and Core Concept Check<\/p>\n

Question 1.
\nVOCABULARY<\/strong>
\nThe sine raio compares the length of ______________ to the length of _____________
\nAnswer:
\n\"Big<\/p>\n

Question 2.
\nWHICH ONE DOESN\u2019T BELONG?<\/strong>
\nWhich ratio does not belong with the other three? Explain your reasoning.
\n\"Big
\nsin B<\/p>\n

Answer:
\nsin B = \\(\\frac { AC }{ BC } \\)<\/p>\n

cos C<\/p>\n

Answer:
\ncos C = \\(\\frac { AC }{ BC } \\)<\/p>\n

tan B<\/p>\n

Answer:
\ntan B = \\(\\frac { AC }{ AB } \\)<\/p>\n\\(\\frac{A C}{B C}\\)\n

Answer:
\n\\(\\frac{A C}{B C}\\) = sin B<\/p>\n

Monitoring Progress and Modeling with Mathematics<\/p>\n

In Exercises 3 – 8, find sin D, sin E, cos D, and cos E. Write each answer as a Fraction and as a decimal rounded to four places.<\/p>\n

Question 3.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 4.
\n\"Big<\/p>\n

Answer:
\nsin D = \\(\\frac { 35 }{ 37 } \\)
\nsin E = \\(\\frac { 12 }{ 37 } \\)
\ncos D = \\(\\frac { 12 }{ 37 } \\)
\ncos E = \\(\\frac { 35 }{ 37 } \\)<\/p>\n

Question 5.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 6.
\n\"Big<\/p>\n

Answer:
\nsin D = \\(\\frac { 36 }{ 45 } \\)
\nsin E = \\(\\frac { 27 }{ 45 } \\)
\ncos D = \\(\\frac { 27 }{ 45 } \\)
\ncos E = \\(\\frac { 36 }{ 45 } \\)<\/p>\n

Question 7.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 8.
\n\"Big<\/p>\n

Answer:
\nsin D = \\(\\frac { 8 }{ 17 } \\)
\nsin E = \\(\\frac { 15 }{ 17 } \\)
\ncos D = \\(\\frac { 15 }{ 17 } \\)
\ncos E = \\(\\frac { 8 }{ 17 } \\)<\/p>\n

In Exercises 9 – 12. write the expression in terms of cosine.<\/p>\n

Question 9.
\nsin 37\u00b0
\nAnswer:
\n\"Big<\/p>\n

Question 10.
\nsin 81\u00b0<\/p>\n

Answer:
\nsin 81\u00b0 = cos(90\u00b0 – 81\u00b0) = cos9\u00b0<\/p>\n

Question 11.
\nsin 29\u00b0
\nAnswer:
\n\"Big<\/p>\n

Question 12.
\nsin 64\u00b0<\/p>\n

Answer:
\nsin 64\u00b0 = cos(90\u00b0 – 64\u00b0) = cos 26\u00b0<\/p>\n

In Exercise 13 – 16, write the expression in terms of sine.<\/p>\n

Question 13.
\ncos 59\u00b0
\nAnswer:
\n\"Big<\/p>\n

Question 14.
\ncos 42\u00b0<\/p>\n

Answer:
\ncos 42\u00b0 = sin(90\u00b0 – 42\u00b0) = sin 48\u00b0<\/p>\n

Question 15.
\ncos 73\u00b0
\nAnswer:
\n\"Big<\/p>\n

Question 16.
\ncos 18\u00b0<\/p>\n

Answer:
\ncos 18\u00b0 = sin(90\u00b0 – 18\u00b0) = sin 72\u00b0<\/p>\n

In Exercises 17 – 22, find the value of each variable using sine and cosine. Round your answers to the nearest tenth.<\/p>\n

Question 17.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 18.
\n\"Big<\/p>\n

Answer:
\np = 30.5, q = 14.8<\/p>\n

Explanation:
\nsin 64\u00b0 = \\(\\frac { p }{34 } \\)
\np = 0.898 x 34
\np = 30.5
\ncos 64\u00b0 = \\(\\frac { q }{ 34 } \\)
\nq = 0.4383 x 34
\nq = 14.8<\/p>\n

Question 19.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 20.
\n\"Big<\/p>\n

Answer:
\ns = 17.7, r = 19<\/p>\n

Explanation:
\nsin 43\u00b0 = \\(\\frac { s }{26 } \\)
\ns = 0.681 x 26
\ns = 17.7
\ncos 43\u00b0 = \\(\\frac { r }{ 26 } \\)
\nr = 0.731 x 26
\nr = 19<\/p>\n

Question 21.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 22.
\n\"Big<\/p>\n

Answer:
\nm = 6.7, n = 10.44<\/p>\n

Explanation:
\nsin 50\u00b0 = \\(\\frac { 8 }{n } \\)
\n0.766 = \\(\\frac { 8 }{n } \\)
\nn = 10.44
\ncos 50\u00b0 = \\(\\frac { m }{ n } \\)
\n0.642 = \\(\\frac { m }{ 10.44 } \\)
\nm = 6.7<\/p>\n

Question 23.
\nREASONING<\/strong>
\nWhich ratios are equal? Select all that apply.
\n\"Big
\nsin X<\/p>\n

cos X<\/p>\n

sin Z<\/p>\n

cos Z
\nAnswer:
\n\"Big<\/p>\n

Question 24.
\nREASONING<\/strong>
\nWhich ratios arc equal to \\(\\frac{1}{2}\\) Select all
\n\"Big
\nsin L<\/p>\n

Answer:
\nsin L = \\(\\frac { 2 }{ 4 } \\) = \\(\\frac { 1 }{ 2 } \\)<\/p>\n

cos L<\/p>\n

Answer:
\ncos L = \\(\\frac { 2\u221a3 }{ 4 } \\) = \\(\\frac { \u221a3 }{ 2 } \\)<\/p>\n

sin J<\/p>\n

Answer:
\nsin J = \\(\\frac { 2\u221a3 }{ 4 } \\) = \\(\\frac { \u221a3 }{ 2 } \\)<\/p>\n

cos J<\/p>\n

Answer:
\ncos J = \\(\\frac { 2 }{ 4 } \\) = \\(\\frac { 1 }{ 2 } \\)<\/p>\n

Question 25.
\nERROR ANALYSIS<\/strong>
\nDescribe and correct the error in finding sin A.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 26.
\nWRITING<\/strong>
\nExplain how to tell which side of a right triangle is adjacent to an angle and which side is the hypotenuse.
\nAnswer:<\/p>\n

Question 27.
\nMODELING WITH MATHEMATICS<\/strong>
\nThe top of the slide is 12 feet from the ground and has an angle of depression of 53\u00b0. What is the length of the slide?
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 28.
\nMODELING WITH MATHEMATICS<\/strong>
\nFind the horizontal distance x the escalator covers.
\n\"Big<\/p>\n

Answer:
\ncos 41 = \\(\\frac { x }{ 26 } \\)
\n0.754 = \\(\\frac { x }{ 26 } \\)
\nx = 19.6 ft<\/p>\n

Question 29.
\nPROBLEM SOLVING<\/strong>
\nYou are flying a kite with 20 feet of string extended. The angle of elevation from the spool of string to the kite is 67\u00b0.
\na. Draw and label a diagram that represents the situation.
\nb. How far off the ground is the kite if you hold the spool 5 feet off the ground? Describe how the height where you hold the spool affects the height of the kite.
\nAnswer:
\n\"Big<\/p>\n

Question 30.
\nMODELING WITH MATHEMATICS<\/strong>
\nPlanes that fly at high speeds and low elevations have radar s sterns that can determine the range of an obstacle and the angle of elevation to the top of the obstacle. The radar of a plane flying at an altitude of 20,000 feet detects a tower that is 25,000 feet away. with an angle of elevation of 1\u00b0
\n\"Big
\na. How many feet must the plane rise to pass over the tower?<\/p>\n

Answer:
\nsin 1 = \\(\\frac { h }{ 25000 } \\)
\n0.017 = \\(\\frac { h }{ 25000 } \\)
\nh = 425 ft
\n425 ft the plane rise to pass over the tower<\/p>\n

b. PIanes Caillot come closer than 1000 feet vertically to any object. At what altitude must the plane fly in order to pass over the tower?<\/p>\n

Answer:<\/p>\n

Question 31.
\nMAKING AN ARGUMENT<\/strong>
\nYour friend uses che equation sin 49\u00b0 = \\(\\frac{x}{16}\\) to find BC. Your cousin uses the equation cos 41\u00b0 = \\(\\frac{x}{16}\\) to find BC. Who is correct? Explain your reasoning.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 32.
\nWRITING<\/strong>
\nDescribe what you must know about a triangle in order to use the sine ratio and what you must know about a triangle in order to use the cosine ratio<\/p>\n

Answer:
\nsin = \\(\\frac { opposite side }{ hypotenuse } \\)
\ncos = \\(\\frac { adjacent side }{ hypotenuse } \\)<\/p>\n

Question 33.
\nMATHEMATICAL CONNECTIONS<\/strong>
\nIf \u2206EQU is equilateral and \u2206RGT is a right triangle with RG = 2, RT = 1. and m \u2220 T = 90\u00b0, show that sin E = cos G.
\nAnswer:
\n\"Big<\/p>\n

Question 34.
\nMODELING WITH MATHEMATICS<\/strong>
\nSubmarines use sonar systems, which are similar to radar systems, to detect obstacles, Sonar systems use sound to detect objects under water.
\n\"Big<\/p>\n

a. You are traveling underwater in a submarine. The sonar system detects an iceberg 4000 meters a head, with an angle of depression of 34\u00b0 to the bottom of the iceberg. How many meters must the submarine lower to pass under the iceberg?<\/p>\n

Answer:
\ntan 34 = \\(\\frac { x }{ 4000 } \\)
\n.674 = \\(\\frac { x }{ 4000 } \\)
\nx = 2696<\/p>\n

b. The sonar system then detects a sunken ship 1500 meters ahead. with an angle of elevation of 19\u00b0 to the highest part of the sunken ship. How many meters must the submarine rise to pass over the sunken ship?<\/p>\n

Answer:
\ntan 19 = \\(\\frac { x }{ 1500 } \\)
\n0.344 = \\(\\frac { x }{ 1500 } \\)
\nx = 516 m<\/p>\n

Question 35.
\nABSTRACT REASONING<\/strong>
\nMake a conjecture about how you could use trigonometric ratios to find angle measures in a triangle.
\nAnswer:
\n\"Big<\/p>\n

Question 36.
\nHOW DO YOU SEE IT?<\/strong>
\nUsing only the given information, would you use a sine ratio or a cosine ratio to find the length of the hypotenuse? Explain your reasoning.
\n\"Big<\/p>\n

Answer:
\nsin 29 = \\(\\frac { 9 }{ x } \\)
\n0.48 = \\(\\frac { 9 }{ x } \\)
\nx = 18.75
\nThe length of hypotenuse is 18.75<\/p>\n

Question 37.
\nMULTIPLE REPRESENTATIONS<\/strong>
\nYou are standing on a cliff above an ocean. You see a sailboat from your vantage point 30 feet above the ocean.<\/p>\n

a. Draw and label a diagram of the situation.
\nb. Make a table showing the angle of depression and the length of your line of sight. Use the angles 40\u00b0, 50\u00b0, 60\u00b0, 70\u00b0, and 80\u00b0.
\nc. Graph the values you found in part (b), with the angle measures on the x-axis.
\nd. Predict the length of the line of sight when the angle of depression is 30\u00b0.
\nAnswer:
\n\"Big
\n\"Big
\n\"Big<\/p>\n

Question 38.
\nTHOUGHT PROVOKING<\/strong>
\nOne of the following infinite series represents sin x and the other one represents cos x (where x is measured in radians). Which is which? Justify your answer. Then use each series to approximate the sine and cosine of \\(\\frac{\\pi}{6}\\).
\n(Hints: \u03c0 = 180\u00b0; 5! = 5 \u2022 4 \u2022 3 \u2022 2 \u2022 1; Find the values that the sine and cosine ratios approach as the angle measure approaches zero).
\na.
\n\"Big<\/p>\n

Answer:
\nFor x = 0
\n0 – \\(\\frac { 0\u00b3 }{ 3! } \\) + \\(\\frac { 0\u2075 }{ 5! } \\) – \\(\\frac { 0\u2077 }{ 7! } \\) + . . . = 0
\nsin x = x – \\(\\frac { x\u00b3 }{ 3! } \\) + \\(\\frac { x\u2075 }{ 5! } \\) – \\(\\frac { x\u2077 }{ 7! } \\) + . . .
\nsin \\(\\frac { \u03c0 }{ 6 } \\) = 0.5<\/p>\n

b.
\n\"Big<\/p>\n

Answer:
\n1 – \\(\\frac { 1\u00b2 }{ 2! } \\) + \\(\\frac { 1\u2074 }{ 4! } \\) – \\(\\frac { 1\u2076 }{ 6! } \\) + . .\u00a0 = 1
\ncos x =x1 – \\(\\frac { x\u00b2 }{ 2! } \\) + \\(\\frac { x\u2074 }{ 4! } \\) – \\(\\frac { x\u2076 }{ 6! } \\) + . .
\ncos \\(\\frac { \u03c0 }{ 6 } \\) = 0.86<\/p>\n

Question 39.
\nCRITICAL THINKING<\/strong>
\nLet A be any acute angle of a right triangle. Show that
\n(a) tan A = \\(\\frac{\\sin A}{\\cos A}\\) and
\n(b) (sin A)2<\/sup> + (cos A)2<\/sup> = 1.
\nAnswer:
\n\"Big<\/p>\n

Question 40.
\nCRITICAL THINKING<\/strong>
\nExplain why the area \u2206 ABC in the diagram can be found using the formula Area = \\(\\frac{1}{2}\\) ab sin C. Then calculate the area when a = 4, b = 7, and m\u2220C = 40\u00b0:
\n\"Big
\nAnswer:
\nArea = \\(\\frac{1}{2}\\) ab sin C
\n= \\(\\frac{1}{2}\\) (4 x 7) sin 40\u00b0
\n= 14 x 0.642
\n= 8.988<\/p>\n

Maintaining Mathematical Proficiency<\/p>\n

Find the value of x. Tell whether the side lengths form a Pythagorean triple.<\/p>\n

Question 41.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 42.
\n\"Big
\nAnswer:
\nx = 12\u221a2<\/p>\n

Explanation:
\nc\u00b2 = a\u00b2 + b\u00b2
\nx\u00b2 = 12\u00b2 + 12\u00b2
\nx\u00b2 = 144 + 144
\nx\u00b2 = 288
\nx = 12\u221a2<\/p>\n

Question 43.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 44.
\n\"Big<\/p>\n

Answer:
\nx = 6\u221a2<\/p>\n

Explanation:
\nc\u00b2 = a\u00b2 + b\u00b2
\n9\u00b2 = x\u00b2 + 3\u00b2
\n81 = x\u00b2 + 9
\nx\u00b2 = 81 – 9
\nx = 6\u221a2<\/p>\n

9.6 Solving Right Triangles<\/h3>\n

Exploration 1<\/strong><\/p>\n

Solving Special Right Triangles<\/p>\n

Work with a partner. Use the figures to find the values of the sine and cosine of \u2220A and \u2220B. Use these values to find the measures of \u2220A and \u2220B. Use dynamic geometry software to verify your answers.
\na.
\n\"Big
\nAnswer:<\/p>\n

b.
\n\"Big
\nAnswer:<\/p>\n

Exploration 2<\/strong><\/p>\n

Solving Right Triangles<\/p>\n

Work with a partner: You can use a calculator to find the measure of an angle when you know the value of the sine, cosine, or tangent of the rule. Use the inverse sine, inverse cosine, 0r inverse tangent feature of your calculator to approximate the measures of \u2220A and \u2220B to the nearest tenth of a degree. Then use dynamic geometry software to verify your answers.
\nATTENDING TO PRECISION<\/strong>
\nTo be proficient in math, you need to calculate accurately and efficiently, expressing numerical answers with a degree of precision appropriate for the problem context.
\na.
\n\"Big
\nAnswer:<\/p>\n

b.
\n\"Big
\nAnswer:<\/p>\n

Communicate Your Answer<\/p>\n

Question 3.
\nWhen you know the lengths of the sides of a right triangle, how can you find the measures of the two acute angles?
\nAnswer:<\/p>\n

Question 4.
\nA ladder leaning against a building forms a right triangle with the building and the ground. The legs of the right triangle (in meters) form a 5-12-13 Pythagorean triple. Find the measures of the two acute angles to the nearest tenth of a degree.
\nAnswer:<\/p>\n

Lesson 9.6 Solving Right Triangles<\/h3>\n

Determine which of the two acute angles has the given trigonometric ratio.<\/p>\n

\"Big<\/p>\n

Question 1.
\nThe sine of the angle is \\(\\frac{12}{13}\\).<\/p>\n

Answer:
\nSin E = \\(\\frac{12}{13}\\)
\nm\u2220E = sin-1<\/sup>(\\(\\frac{12}{13}\\)) = 67.3\u00b0<\/p>\n

Question 2.
\nThe tangent of the angle is \\(\\frac{5}{12}\\)<\/p>\n

Answer:
\ntan F = \\(\\frac{5}{12}\\)
\nm\u2220F = tan-1<\/sup>(\\(\\frac{5}{12}\\)) = 22.6\u00b0<\/p>\n

Let \u2220G, \u2220H, and \u2220K be acute angles. Use a calculator to approximate the measures of \u2220G, \u2220H, and \u2220K to the nearest tenth of a degree.<\/p>\n

Question 3.
\ntan G = 0.43<\/p>\n

Answer:
\n\u2220G = inverse tan of 0.43 = 23.3\u00b0<\/p>\n

Question 4.
\nsin H = 0.68<\/p>\n

Answer:
\n\u2220H = inverse sin of 0.68 = 42.8\u00b0<\/p>\n

Question 5.
\ncos K = 0.94<\/p>\n

Answer:
\n\u2220K = inverse cos of 0.94 = 19.9\u00b0<\/p>\n

Solve the right triangle. Round decimal answers to the nearest tenth.<\/p>\n

Question 6.
\n\"Big<\/p>\n

Answer:
\nDE = 29, \u2220D = 46.05\u00b0, \u2220E = 42.84\u00b0<\/p>\n

Explanation:
\nc\u00b2 = a\u00b2 + b\u00b2
\nx\u00b2 = 20\u00b2 + 21\u00b2
\nx\u00b2 = 400 + 441
\nx\u00b2 = 841
\nx = 29
\nsin D = \\(\\frac { 21 }{ 29 } \\)
\n\u2220D = 46.05
\nsin E = \\(\\frac { 20 }{ 29 } \\)
\n\u2220E = 42.84<\/p>\n

Question 7.
\n\"Big<\/p>\n

Answer:
\nGJ = 60, \u2220G = 56.09\u00b0, \u2220H = 33.3\u00b0<\/p>\n

Explanation:
\nc\u00b2 = a\u00b2 + b\u00b2
\n109\u00b2 = 91\u00b2 + x\u00b2
\nx\u00b2 = 11881 – 8281
\nx\u00b2 = 3600
\nx = 60
\nsin G = \\(\\frac { 91 }{ 109 } \\)
\n\u2220G = 56.09
\nsin H = \\(\\frac { 60 }{ 109 } \\)
\n\u2220H = 33.3<\/p>\n

Question 8.
\nSolve the right triangle. Round decimal answers to the nearest tenth.
\n\"Big<\/p>\n

Answer:
\nXY = 13.82, YZ = 6.69, \u2220Y = 37.5<\/p>\n

Explanation:
\ncos 52 = \\(\\frac { 8.5 }{ XY } \\)
\n0.615 = \\(\\frac { 8.5 }{ XY } \\)
\nXY = 13.82
\nsin 52 = \\(\\frac { YZ }{ XY } \\)
\n0.788 = \\(\\frac { YZ }{ 8.5 } \\)
\nYZ = 6.69
\nsin Y = \\(\\frac { 8.5 }{ 13.82 } \\)
\n\u2220Y = 37.5<\/p>\n

Question 9.
\nWHAT IF?<\/strong>
\nIn Example 5, suppose another raked stage is 20 feet long from front to back with a total rise of 2 feet. Is the raked stage within your desired range?<\/p>\n

Answer:
\nx = inverse sine of \\(\\frac { 2 }{ 20 } \\)
\nx = 5.7\u00b0<\/p>\n

Exercise 9.6 Solving Right Triangles<\/h3>\n

Question 1.
\nCOMPLETE THE SENTENCE<\/strong>
\nTo solve a right triangle means to find the measures of all its ________ and _______ .
\nAnswer:
\n\"Big<\/p>\n

Question 2.
\nWRITING<\/strong>
\nExplain when you can use a trigonometric ratio to find a side length of a right triangle and when you can use the Pythagorean Theorem (Theorem 9.1 ).
\nAnswer:<\/p>\n

Monitoring Progress and Modeling with Mathematics<\/p>\n

In Exercises 3 – 6. determine which of the two acute angles has the given trigonometric ratio.<\/p>\n

Question 3.
\nThe cosine of the angle is \\(\\frac{4}{5}\\)
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 4.
\nThe sine of the angle is \\(\\frac{5}{11}\\)
\n\"Big<\/p>\n

Answer:
\nSin(angle) = \\(\\frac { opposite }{ hypo } \\)
\nsin A = \\(\\frac{5}{11}\\)
\nThe acute angle that has a sine of the angle is \\(\\frac{5}{11}\\) is \u2220A.<\/p>\n

Question 5.
\nThe sine of the angle is 0.95.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 6.
\nThe tangent of the angle is 1.5.
\n\"Big<\/p>\n

Answer:
\ntan(angle) = \\(\\frac { opposite }{ adjacent } \\)
\n1.5 = \\(\\frac { 18 }{ 12 } \\)
\ntan C = 1.5
\nThe acute angle that has a tangent of the angle is 1.5 \u2220C.<\/p>\n

In Exercises 7 – 12, let \u2220D be an acute angle. Use a calculator to approximate the measure of \u2220D to the nearest tenth of a degree.<\/p>\n

Question 7.
\nsin D = 0.75
\nAnswer:
\n\"Big<\/p>\n

Question 8.
\nsin D = 0.19<\/p>\n

Answer:
\nsin D = 0.19
\n\u2220D = inverse sine of 0.19
\n\u2220D = 10.9\u00b0<\/p>\n

Question 9.
\ncos D = 0.33
\nAnswer:
\n\"Big<\/p>\n

Question 10.
\ncos D = 0.64<\/p>\n

Answer:
\ncos D = 0.64
\n\u2220D = inverse cos of 0.64
\n\u2220D = 50.2\u00b0<\/p>\n

Question 11.
\ntan D = 0.28
\nAnswer:
\n\"Big<\/p>\n

Question 12.
\ntan D = 0.72<\/p>\n

Answer:
\ntan D = 0.72
\n\u2220D = inverse tan of 0.72
\n\u2220D = 35.8\u00b0<\/p>\n

In Exercises 13 – 18. solve the right triangle. Round decimal answers to the nearest tenth.<\/p>\n

Question 13.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 14.
\n\"Big
\nAnswer:
\nED = 2\u221a65, \u2220E = 59.3, \u2220D = 29.7<\/p>\n

Explanation:
\nc\u00b2 = 8\u00b2 + 14\u00b2
\nx\u00b2 = 64 + 196
\nx\u00b2 = 260
\nx = 2\u221a65
\nsin E = \\(\\frac { 14 }{ 2\u221a65 } \\)
\n\u2220E = 59.3
\nsin D = \\(\\frac { 8 }{ 2\u221a65 } \\)
\n\u2220D = 29.7<\/p>\n

Question 15.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 16.
\n\"Big
\nAnswer:
\nHJ = 2\u221a15, \u2220G = 28.9, \u2220J = 61<\/p>\n

Explanation:
\nc\u00b2 = a\u00b2 + b\u00b2
\n16\u00b2 = 14\u00b2 + x\u00b2
\nx\u00b2 = 256 – 196
\nx\u00b2 = 60
\nx = 2\u221a15
\nsin G = \\(\\frac { 2\u221a15 }{ 16 } \\)
\n\u2220G = 28.9
\nsin J = \\(\\frac { 14 }{ 16 } \\)
\n\u2220J = 61<\/p>\n

Question 17.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 18.
\n\"Big<\/p>\n

Answer:
\nRT = 17.8, RS = 9.68, \u2220T = 32.8<\/p>\n

Explanation:
\nsin 57 = \\(\\frac { 15 }{ x } \\)
\n0.838 = \\(\\frac { 15 }{ x } \\)
\nx = 17.899
\nRT = 17.8
\ncos 57 = \\(\\frac { x }{ 17.8 } \\)
\n0.544 = \\(\\frac { x }{ 17.8 } \\)
\nx = 9.68
\nRS = 9.68
\nsin T = \\(\\frac { 9.68 }{ 17.8 } \\)
\n\u2220T = 32.8<\/p>\n

Question 19.
\nERROR ANALYSIS<\/strong>
\nDescribe and correct the error in using an inverse trigonometric ratio.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 20.
\nPROBLEM SOLVING<\/strong>
\nIn order to unload clay easily. the body of a dump truck must be elevated to at least 45\u00b0 The body of a dump truck that is 14 feet long has been raised 8 feet. Will the clay pour out easily? Explain your reasoning.<\/p>\n

Answer:
\nAngle of elevation: sin x = \\(\\frac { 8 }{ 14 } \\)
\nx = inverse sine of \\(\\frac { 8 }{ 14 } \\) = 34.9
\nThe clay will not pour out easily.<\/p>\n

Question 21.
\nPROBLEM SOLVING<\/strong>
\nYou are standing on a footbridge that is 12 feet above a lake. You look down and see a duck in the water. The duck is 7 feet away from the footbridge. What is the angle of elevation from the duck to you
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 22.
\nHOW DO YOU SEE IT?<\/strong>
\nWrite three expressions that can be used to approximate the measure of \u2220A. Which expression would you choose? Explain your choice.
\n\"Big<\/p>\n

Answer:
\nThree expressions are \u2220A = inverse tan of (\\(\\frac { 15 }{ 22 } \\)) = 34.2\u00b0
\n\u2220A = inverse sine of (\\(\\frac { 15 }{ BA } \\))
\n\u2220A = inverse cos of (\\(\\frac { 22 }{ BA } \\))<\/p>\n

Question 23.
\nMODELING WITH MATHEMATICS<\/strong>
\nThe Uniform Federal Accessibility Standards specify that awheel chair ramp may not have an incline greater than 4.76. You want to build a ramp with a vertical rise of 8 inches. you want to minimize the horizontal distance taken up by the ramp. Draw a diagram showing the approximate dimensions of your ramp.
\nAnswer:
\n\"Big<\/p>\n

Question 24.
\nMODELING WITH MATHEMATICS<\/strong>
\nThe horizontal part of a step is called the tread. The vertical part is called the riser. The recommended riser – to – tread ratio is 7 inches : 11 inches.<\/p>\n

a. Find the value of x for stairs built using the recommended riser-to-tread ratio.
\n\"Big<\/p>\n

Answer:<\/p>\n

b. you want to build stairs that are less steep than the stairs in part (a). Give an example of a riser – to – tread ratio that you could use. Find the value of x for your stairs.
\nAnswer:<\/p>\n

Question 25.
\nUSING TOOLS<\/strong>
\nFind the measure of \u2220R without using a protractor. Justify your technique.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 26.
\nMAKING AN ARGUMENT<\/strong>
\nYour friend claims that tan-1<\/sup>x = \\(\\frac{1}{\\tan x}\\). Is your friend correct? Explain your reasoning.<\/p>\n

Answer:
\nNo
\nFor example
\ntan-1<\/sup>(\u221a3) = 60
\n\\(\\frac{1}{\\tan \u221a3}\\) = 33.1<\/p>\n

USING STRUCTURE<\/strong>
\nIn Exercises 27 and 28, solve each triangle.<\/p>\n

Question 27.
\n\u2206JKM and \u2206LKM
\n\"Big
\nAnswer:
\n\"Big
\n\"Big<\/p>\n

Question 28.
\n\u2206TUS and \u2206VTW
\n\"Big<\/p>\n

Answer:
\nTS = 8.2, UT = 7.3, \u2220T = 28.6
\nTV = 13.2, TW = 9.6, \u2220V = 46, \u2220T = 42.84<\/p>\n

Explanation:
\ntan 64 = \\(\\frac { TS }{ 4 } \\)
\nTS = 2.05 x 4
\nTS = 8.2
\nsin 64 = \\(\\frac { UT }{ 8.2 } \\)
\n0.898= \\(\\frac { UT }{ 8.2 } \\)
\nUT = 7.3
\nsin T = \\(\\frac { 4 }{ 8.2 } \\)
\n\u2220T = 28.6
\nTV = TS + SV
\nTV = 8.2 + 5 = 13.2
\n13.2\u00b2 = TW\u00b2 + 9\u00b2
\nTW\u00b2 = 174.24 – 81
\nTW = 9.6
\nsin V = \\(\\frac { 9.6 }{ 13.2 } \\)
\n\u2220V = 46
\nsin T = \\(\\frac { 9 }{ 13.2 } \\)
\n\u2220T = 42.84<\/p>\n

Question 29.
\nMATHEMATICAL CONNECTIONS<\/strong>
\nWrite an expression that can be used to find the measure of the acute angle formed by each line and the x-axis. Then approximate the angle measure to the nearest tenth of a degree.
\na. y = 3x
\nb. y = \\(\\frac{4}{3}\\)x + 4
\nAnswer:
\n\"Big<\/p>\n

Question 30.
\nTHOUGHT PROVOKING<\/strong>
\nSimplify each expression. Justify your answer.
\na. sin-1<\/sup> (sin x)<\/p>\n

Answer:
\nsin-1<\/sup> (sin x) = x<\/p>\n

b. tan(tan-1<\/sup> y)<\/p>\n

Answer:
\ntan(tan-1<\/sup> y) = y<\/p>\n

C. cos(cos-1<\/sup> z)<\/p>\n

Answer:
\ncos(cos-1<\/sup> z) = z<\/p>\n

Question 31.
\nREASONING<\/strong>
\nExplain why the expression sin-1<\/sup> (1.2) does not make sense.
\nAnswer:
\n\"Big<\/p>\n

Question 32.
\nUSING STRUCTURE<\/strong>
\nThe perimeter of the rectangle ABCD is 16 centimeters. and the ratio of its width to its length is 1 : 3. Segment BD divides the rectangle into two congruent triangles. Find the side lengths and angle measures of these two triangles.<\/p>\n

Answer:
\nThe perimeter of the rectangle ABCD is 16 centimeters
\n2(l + b) = 16
\nl + b = 8
\nb : l = 1 : 3
\n4x = 8
\nx = 2
\nl = 6, b = 2
\nBD = \u221a(6\u00b2 + 2\u00b2) = \u221a40 = 2\u221a10
\nsin B = \\(\\frac { AD }{ BD } \\) = \\(\\frac { 2 }{ 2\u221a10 } \\)
\n\u2220ABD = 18.4
\n\u2220CBD = 71.6
\nsin D = \\(\\frac { AB }{ BD } \\) = \\(\\frac { 6 }{ 2\u221a10 } \\)
\n\u2220ADB = 71
\n\u2220CDB = 19<\/p>\n

Maintaining Mathematical Proficiency<\/p>\n

Solve the equation<\/p>\n

Question 33.
\n\\(\\frac{12}{x}=\\frac{3}{2}\\)
\nAnswer:
\n\"Big<\/p>\n

Question 34.
\n\\(\\frac{13}{9}=\\frac{x}{18}\\)<\/p>\n

Answer:
\n\\(\\frac{13}{9}=\\frac{x}{18}\\)
\nx = 1.44 x 18
\nx = 26<\/p>\n

Question 35.
\n\\(\\frac{x}{2.1}=\\frac{4.1}{3.5}\\)
\nAnswer:
\n\"Big<\/p>\n

Question 36.
\n\\(\\frac{5.6}{12.7}=\\frac{4.9}{x}\\)<\/p>\n

Answer:
\n\\(\\frac{5.6}{12.7}=\\frac{4.9}{x}\\)
\n0.44 = 4.9\/x
\nx = 11.13<\/p>\n

9.7 Law of Sines and Law of Cosines<\/h3>\n

Exploration 1<\/strong><\/p>\n

Discovering the Law of Sines<\/p>\n

Work with a partner.<\/p>\n

a. Copy and complete the table for the triangle shown. What can you conclude?
\n\"Big
\n\"Big
\nAnswer:<\/p>\n

b. Use dynamic geometry software to draw two other triangles. Copy and complete the table in part (a) for each triangle. Use your results to write a conjecture about the relationship between the sines of the angles and the lengths of the sides of a triangle.
\nUSING TOOLS STRATEGICALLY<\/strong>
\nTo be proficient in math, you need to use technology to compare predictions with data.
\nAnswer:<\/p>\n

Exploration 2<\/strong><\/p>\n

Discovering the Law of Cosines<\/p>\n

Work with a partner:<\/p>\n

a. Copy and complete the table for the triangle in Exploration 1 (a). What can you conclude?
\n\"Big
\nAnswer:<\/p>\n

b. Use dynamic geometry software to draw two other triangles. Copy and complete the table in part (a) for each triangle. Use your results to write a conjecture about what you observe in the completed tables.
\nAnswer:<\/p>\n

Communicate Your Answer<\/p>\n

Question 3.
\nWhat are the Law of Sines and the Law of Cosines?
\nAnswer:<\/p>\n

Question 4.
\nWhen would you use the Law of Sines to solve a triangle? When would you use the Law of Cosines to solve a triangle?
\nAnswer:<\/p>\n

Lesson 9.7 Law of Sines and Law of Cosines<\/h3>\n

Monitoring Progress<\/strong><\/p>\n

Use a calculator to find the trigonometric ratio. Round your answer to four decimal places.<\/p>\n

Question 1.
\ntan 110\u00b0<\/p>\n

Answer:
\ntan 110\u00b0 = -2.7474<\/p>\n

Question 2.
\nsin 97\u00b0<\/p>\n

Answer:
\nsin 97\u00b0 = 0.9925<\/p>\n

Question 3.
\ncos 165\u00b0<\/p>\n

Answer:
\ncos 165\u00b0 = -0.9659<\/p>\n

Find the area of \u2206ABC with the given side lengths and included angle. Round your answer to the nearest tenth.<\/p>\n

Question 4.
\nm \u2220 B = 60\u00b0, a = 19, c = 14<\/p>\n

Answer:
\nArea = 155.18<\/p>\n

Explanation:
\nArea = \\(\\frac { 1 }{ 2 } \\) ac sin B
\n= \\(\\frac { 1 }{ 2 } \\) (19 x 14) sin 60\u00b0
\n= 133 x 0.866
\n= 155.18<\/p>\n

Question 5.
\nm \u2220 C = 29\u00b0, a = 38, b = 31<\/p>\n

Answer:
\nArea = 282.72<\/p>\n

Explanation:
\nArea = \\(\\frac { 1 }{ 2 } \\) ab sin C
\n= \\(\\frac { 1 }{ 2 } \\) (38 x 31) sin 29
\n= 598 x 0.48
\n= 282.72<\/p>\n

Solve the triangle. Round decimal answers to the nearest tenth.<\/p>\n

Question 6.
\n\"Big<\/p>\n

Answer:
\n\u2220C = 46.6, \u2220B = 82.4, AC = 23.57<\/p>\n

Explanation:
\nUsing law of sines
\n\\(\\frac { c }{ sin C } \\) = \\(\\frac { a }{ sin A } \\)
\n\\(\\frac { 17 }{ sin C } \\) = \\(\\frac { 18 }{ sin 51 } \\)
\n\\(\\frac { 17 }{ sin C } \\) = \\(\\frac { 18 }{ 0.77 } \\)
\nsin C = 0.7274
\n\u2220C = 46.6
\n\u2220A + \u2220B + \u2220C\u00a0 = 180
\n51 + 46.6 + \u2220B = 180
\n\u2220B = 82.4
\n\\(\\frac { sin B }{ b } \\) = \\(\\frac { sin A }{ a } \\)
\n\\(\\frac { 0.99 }{ b } \\) = \\(\\frac { 0.77 }{ 18 } \\)
\nb = 23.57<\/p>\n

Question 7.
\n\"Big<\/p>\n

Answer:
\n\u2220B = 31.3, \u2220C\u00a0 = 108.7, c = 23.6<\/p>\n

Explanation:
\n\\(\\frac { sin B }{ b } \\) = \\(\\frac { sin A }{ a } \\)
\n\\(\\frac { sin B }{ 13 } \\) = \\(\\frac { sin 40 }{ 16 } \\)
\nsin B = \\(\\frac { .64 }{ 16 } \\) x 13
\nsin B = 0.52
\n\u2220B = 31.3
\n\u2220A + \u2220B + \u2220C\u00a0 = 180
\n40 + 31.3 + \u2220C\u00a0 = 180
\n\u2220C\u00a0 = 108.7
\n\\(\\frac { c }{ sin C } \\) = \\(\\frac { a }{ sin A } \\)
\n\\(\\frac { c }{ sin 108.7 } \\) = \\(\\frac {16 }{ sin 40 } \\)
\nc = \\(\\frac {16 }{ .64 } \\) x 0.947
\nc = 23.6<\/p>\n

Solve the triangle. Round decimal answers to the nearest tenth.<\/p>\n

Question 8.
\n\"Big<\/p>\n

Answer:
\n\u2220C = 66, a = 4.36, c = 8.27<\/p>\n

Explanation:
\n\u2220A + \u2220B + \u2220C = 180
\n29 + 85 + \u2220C = 180
\n\u2220C = 66
\n\\(\\frac { a }{ sin A } \\) = \\(\\frac { b }{ sin B } \\) = \\(\\frac { c }{ sin C } \\)
\n\\(\\frac { a }{ sin 29 } \\) = \\(\\frac { 9 }{ sin 85 } \\)
\n\\(\\frac { a }{ 0.48 } \\) = \\(\\frac { 9 }{ 0.99 } \\)
\na = 4.36
\n\\(\\frac { b }{ sin B } \\) = \\(\\frac { c }{ sin C } \\)
\n\\(\\frac { 9 }{ sin 85 } \\) = \\(\\frac { c }{ sin 66 } \\)
\n\\(\\frac { 9 }{ 0.99 } \\) = \\(\\frac { c }{ 0.91 } \\)
\nc = 8.27<\/p>\n

Question 9.
\n\"Big<\/p>\n

Answer:
\n\u2220A = 29, b = 19.37, c = 20.41<\/p>\n

Explanation:
\n\u2220A + \u2220B + \u2220C\u00a0 = 180
\n\u2220A + 70 + 81 = 180
\n\u2220A = 29
\n\\(\\frac { a }{ sin A } \\) = \\(\\frac { b }{ sin B } \\) = \\(\\frac { c }{ sin C } \\)
\n\\(\\frac { 10 }{ sin 29 } \\) = \\(\\frac { b }{ sin 70 } \\)
\n\\(\\frac { 10 }{ 0.48 } \\) = \\(\\frac { b }{ 0.93 } \\)
\nb = 19.37
\n\\(\\frac { a }{ sin A } \\) = \\(\\frac { c }{ sin C } \\)
\n\\(\\frac { 10 }{ sin 29 } \\) = \\(\\frac { c }{ sin 81 } \\)
\n\\(\\frac { 10 }{ 0.48 } \\) = \\(\\frac { c }{ 0.98 } \\)
\nc = 20.41<\/p>\n

Question 10.
\nWHAT IF?<\/strong>
\nIn Example 5, what would be the length of a bridge from the South Picnic Area to the East Picnic Area?<\/p>\n

Answer:
\nThe length of a bridge from the South Picnic Area to the East Picnic Area is 188 m.<\/p>\n

Explanation:
\n\\(\\frac { a }{ sin A } \\) = \\(\\frac { b }{ sin B } \\) = \\(\\frac { c }{ sin C } \\)
\n\\(\\frac { a }{ sin 71 } \\) = \\(\\frac { 150 }{ sin 49 } \\)
\na = 188<\/p>\n

Solve the triangle. Round decimal answers to the nearest tenth.<\/p>\n

Question 11.
\n\"Big<\/p>\n

Answer:
\nb = 61.3, \u2220A = 46, \u2220C\u00a0 = 46<\/p>\n

Explanation:
\nb\u00b2 = a\u00b2 + c\u00b2 \u2212 2ac cos B
\nb\u00b2 = 45\u00b2 + 43\u00b2 – 2(45)(43) cos 88
\nb\u00b2 = 2025 + 1849 – 3870 x 0.03 = 3757.9
\nb = 61.3
\n\\(\\frac { sin A }{ a } \\) = \\(\\frac { sin B }{ b } \\)
\n\\(\\frac { sin A }{ 45 } \\) = \\(\\frac { sin 88 }{ 61.3 } \\)
\nsin A = 0.72
\n\u2220A = 46
\n\u2220A + \u2220B + \u2220C\u00a0 = 180
\n46 + 88 + \u2220C\u00a0 = 180
\n\u2220C\u00a0 = 46<\/p>\n

Question 12.
\n\"Big<\/p>\n

Answer:
\na = 41.1, \u2220C\u00a0 = 35.6, \u2220B = 30.4<\/p>\n

Explanation:
\na\u00b2 = b\u00b2 + c\u00b2 \u2212 2bc cos A
\na\u00b2 = 23\u00b2 + 26\u00b2 – 2(23)(26) cos 114
\na\u00b2 = 529 + 676 – 1196 x -0.406
\na\u00b2 = 1690.5
\na = 41.1
\n\\(\\frac { sin 114 }{ 41.1 } \\) = \\(\\frac { sin B }{ 23 } \\)
\n0.02 = \\(\\frac { sin B }{ 23 } \\)
\nsin B = 0.507
\n\u2220B = 30.4
\n\u2220A + \u2220B + \u2220C\u00a0 = 180
\n114 + 30.4 + \u2220C\u00a0 = 180
\n\u2220C\u00a0 = 35.6<\/p>\n

Question 13.
\n\"Big
\nAnswer:
\n\u2220A = 41.4, \u2220B = 81.8, \u2220C\u00a0 = 56.8<\/p>\n

Explanation:
\na\u00b2 = b\u00b2 + c\u00b2 \u2212 2bc cos A
\n4\u00b2 = 6\u00b2 + 5\u00b2 – 2(6)(5) cos A
\n16 = 36 + 25 – 60 cos A
\n-45 = – 60 cos A
\ncos A = 0.75
\n\u2220A = 41.4
\n\\(\\frac { sin 41.4 }{ 4 } \\) = \\(\\frac { sin B }{ 6 } \\)
\n0.165 = \\(\\frac { sin B }{ 6 } \\)
\nsin B = 0.99
\n\u2220B = 81.8
\n\u2220A + \u2220B + \u2220C\u00a0 = 180
\n41.4 + 81.8 + \u2220C\u00a0 = 180
\n\u2220C\u00a0 = 56.8<\/p>\n

Question 14.
\n\"Big<\/p>\n

Answer:
\n\u2220B = 81.8, \u2220A = 58.6, \u2220C\u00a0 = 39.6<\/p>\n

Explanation:
\na\u00b2 = b\u00b2 + c\u00b2 \u2212 2bc cos A
\n23\u00b2 = 27\u00b2 + 16\u00b2 – 2(27)(16) cos A
\n529 = 729 + 256 – 864 cos A
\n456 = 864 cos A
\ncos A = 0.52
\n\u2220A = 58.6
\n\\(\\frac { sin 58.6 }{ 23 } \\) = \\(\\frac { sin B }{ 27 } \\)
\n0.03 = \\(\\frac { sin B }{ 27 } \\)
\nsin B = 0.99
\n\u2220B = 81.8
\n\u2220A + \u2220B + \u2220C\u00a0 = 180
\n58.6 + 81.8 + \u2220C\u00a0 = 180
\n\u2220C\u00a0 = 39.6<\/p>\n

Exercise 9.7 Law of Sines and Law of Cosines<\/h3>\n

Vocabulary and Core Concept Check<\/p>\n

Question 1.
\nWRITING<\/strong>
\nWhat type of triangle would you use the Law of Sines or the Law of Cosines to solve?
\nAnswer:
\n\"Big<\/p>\n

Question 2.
\nVOCABULARY<\/strong>
\nWhat information do you need to use the Law of Sines?<\/p>\n

Answer:<\/p>\n

Monitoring progress and Modeling with Mathematics<\/p>\n

In Exercises 3 – 8, use a calculator to find the trigonometric ratio, Round your answer to four decimal places.<\/p>\n

Question 3.
\nsin 127\u00b0
\nAnswer:
\n\"Big<\/p>\n

Question 4.
\nsin 98\u00b0<\/p>\n

Answer:
\nsin 98\u00b0 = 0.9902<\/p>\n

Question 5.
\ncos 139\u00b0
\nAnswer:
\n\"Big<\/p>\n

Question 6.
\ncos 108\u00b0<\/p>\n

Answer:
\ncos 108\u00b0 = -0.309<\/p>\n

Question 7.
\ntan 165\u00b0
\nAnswer:
\n\"Big<\/p>\n

Question 8.
\ntan 116\u00b0<\/p>\n

Answer:
\ntan 116\u00b0 = -2.0503<\/p>\n

In Exercises 9 – 12, find the area of the triangle. Round your answer to the nearest tenth.<\/p>\n

Question 9.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 10.
\n\"Big<\/p>\n

Answer:
\nArea = \\(\\frac{1}{2}\\)bc sin A
\nArea = \\(\\frac{1}{2}\\)(28)(24) sin83
\nArea = 332.64<\/p>\n

Question 11.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 12.
\n\"Big<\/p>\n

Answer:
\nArea = \\(\\frac{1}{2}\\)ab sin C
\nArea = \\(\\frac{1}{2}\\)(15)(7) sin 96
\nArea = 51.9<\/p>\n

In Exercises 13 – 18. solve the triangle. Round decimal answers to the nearest tenth.<\/p>\n

Question 13.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 14.
\n\"Big<\/p>\n

Answer:
\n\u2220B = 38.3, \u2220A = 37.7, a = 15.7<\/p>\n

Explanation:
\n\\(\\frac { sin B }{16 } \\) = \\(\\frac { sin 104 }{ 25 } \\)
\nsin B = 0.62
\n\u2220B = 38.3
\n\u2220A + \u2220B + \u2220C = 180
\n\u2220A + 38.3 + 104 = 180
\n\u2220A = 37.7
\n\\(\\frac { sin 37.7 }{ a } \\) = \\(\\frac { sin 104 }{ 25 } \\)
\n\\(\\frac { 0.61 }{ a } \\) = 0.0388
\na = 15.7<\/p>\n

Question 15.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 16.
\n\"Big<\/p>\n

Answer:
\n\u2220B = 65, b = 33.55, a = 24.4<\/p>\n

Explanation:
\n\u2220A + \u2220B + \u2220C = 180
\n42 + 73 + \u2220B = 180
\n\u2220B = 65
\n\\(\\frac { sin B }{ b } \\) = \\(\\frac { sin C }{ c } \\)
\n\\(\\frac { sin 65 }{ b } \\) = \\(\\frac { sin 73 }{ 34 } \\)
\nb = 33.55
\n\\(\\frac { sin A }{ a } \\) = \\(\\frac { sin C }{ c } \\)
\n\\(\\frac { sin 42 }{ a } \\) = \\(\\frac { sin 73 }{ 34 } \\)
\na = 24.4<\/p>\n

Question 17.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 18.
\n\"Big<\/p>\n

Answer:
\n\u2220C = 90, b = 39.56, a = 17.6<\/p>\n

Explanation:
\n\u2220A + \u2220B + \u2220C = 180
\n24 + 66 + \u2220C = 180
\n\u2220C = 90
\n\\(\\frac { sin B }{ b } \\) = \\(\\frac { sin C }{ c } \\)
\n\\(\\frac { sin 66 }{ b } \\) = \\(\\frac { sin 90 }{ 43 } \\)
\n\\(\\frac { .91 }{ b } \\) = 0.023
\nb = 39.56
\n\\(\\frac { sin A }{ a } \\) = \\(\\frac { sin C }{ c } \\)
\n\\(\\frac { sin 24 }{a } \\) = \\(\\frac { sin 90 }{ 43 } \\)
\n\\(\\frac { 0.406 }{a } \\) = 0.023
\na = 17.6<\/p>\n

In Exercises 19 – 24, solve the triangle. Round decimal answers to the nearest tenth.<\/p>\n

Question 19.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 20.
\n\"Big<\/p>\n

Answer:
\nb = 29.9, \u2220A = 26.1, \u2220C = 15.07<\/p>\n

Explanation:
\nb\u00b2 = a\u00b2 + c\u00b2 – 2ac cos B
\nb\u00b2 = 20\u00b2 + 12\u00b2 – 2(12)(20) cos 138
\nb\u00b2 = 400 + 144 – 480 (-0.74)
\nb\u00b2 = 899.2
\nb = 29.9
\n\\(\\frac { sin B }{ b } \\) = \\(\\frac { sin A }{ a } \\)
\n\\(\\frac { sin 138 }{ 29.9 } \\) = \\(\\frac { sin A }{ 20 } \\)
\n\\(\\frac { 0.66 }{ 29.9 } \\) = \\(\\frac { sin A }{ 20 } \\)
\nsin A = 0.44
\n\u2220A = 26.1
\n\\(\\frac { sin B }{ b } \\) = \\(\\frac { sin C }{ c } \\)
\n\\(\\frac { sin 138 }{ 29.9 } \\) = \\(\\frac { sin C }{ 12 } \\)
\nsin C = 0.26
\n\u2220C = 15.07<\/p>\n

Question 21.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 22.
\n\"Big<\/p>\n

Answer:
\n\u2220A = 107.3, \u2220B = 51.6, \u2220C = 21.1<\/p>\n

Explanation:
\nb\u00b2 = a\u00b2 + c\u00b2 – 2ac cos B
\n28\u00b2 = 18\u00b2 + 13\u00b2 – 2(18)(13) cos B
\n784 = 324 + 169 – 468 cos B
\n291 = 468 cos B
\ncos B = 0.62
\n\u2220B = 51.6
\n\\(\\frac { sin C }{ c } \\) = \\(\\frac { sin B }{ b } \\)
\n\\(\\frac { sin C }{ 13 } \\) = \\(\\frac { sin 51.6 }{ 28 } \\)
\nsin C = 0.36
\n\u2220C = 21.1
\n\u2220A + \u2220B + \u2220C = 180
\n51.6 + 21.1 + \u2220A = 180
\n\u2220A = 107.3<\/p>\n

Question 23.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 24.
\n\"Big<\/p>\n

Answer:
\n\u2220A = 23, \u2220B = 132.1, \u2220C = 24.9<\/p>\n

Explanation:
\nb\u00b2 = a\u00b2 + c\u00b2 – 2ac cos B
\n5\u00b2 = 12\u00b2 + 13\u00b2 – 2(12)(13) cos B
\n25 = 144 + 169 – 312 cos B
\n288 = 312 cos B
\ncos B = 0.92
\n\u2220B = 23
\n\\(\\frac { sin C }{ c } \\) = \\(\\frac { sin B }{ b } \\)
\n\\(\\frac { sin C }{ 13 } \\) = \\(\\frac { sin 23 }{ 5 } \\)
\nsin C = 1.014
\n\u2220C = 24.9
\n\u2220A + \u2220B + \u2220C = 180
\n23 + 24.9 + \u2220B = 180
\n\u2220B = 132.1<\/p>\n

Question 25.
\nERROR ANALYSIS<\/strong>
\nDescribe and correct the error in finding m \u2220 C.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 26.
\nERROR ANALYSIS<\/strong>
\nDescribe and correct the error in finding m \u2220 A in \u2206ABC when a = 19, b = 21, and c = 11.
\n\"Big<\/p>\n

Answer:
\na\u00b2 = b\u00b2 + c\u00b2 – 2bc cos A
\n19\u00b2 = 21\u00b2 + 11\u00b2 – 2(21)(11) cos A
\n361 = 441 + 121 – 462 cosA
\n201 = 462 cosA
\ncos A = 0.43
\n\u2220A = 64.5<\/p>\n

COMPARING METHODS<\/strong><\/p>\n

In Exercise 27 – 32. tell whether you would use the Law of Sines, the Law of Cosines. or the Pythagorean Theorem (Theorem 9.1) and trigonometric ratios to solve the triangle with the given information. Explain your reasoning. Then solve the triangle.<\/p>\n

Question 27.
\nm \u2220 A = 72\u00b0, m \u2220 B = 44\u00b0, b = 14
\nAnswer:
\n\"Big<\/p>\n

Question 28.
\nm \u2220 B = 98\u00b0, m \u2220 C = 37\u00b0, a = 18<\/p>\n

Answer:
\n\u2220A = 45, b = 25.38, c = 15.38<\/p>\n

Explanation:
\n\u2220A + \u2220B + \u2220C = 180
\n\u2220A + 98 + 37 = 180
\n\u2220A = 45
\n\\(\\frac { sin B }{ b } \\) = \\(\\frac { sin A }{ a } \\)
\n\\(\\frac { sin 98 }{ b } \\) = \\(\\frac { sin 45 }{ 18 } \\)
\n\\(\\frac { 0.99 }{ b } \\) = 0.039
\nb = 25.38
\n\\(\\frac { sin A }{ a } \\) = \\(\\frac { sin C }{ c } \\)
\n\\(\\frac { sin 45 }{ 18 } \\) = \\(\\frac { sin 37 }{ c } \\)
\n0.039 = \\(\\frac { sin 37 }{ c } \\)
\nc = 15.38<\/p>\n

Question 29.
\nm \u2220 C = 65\u00b0, a = 12, b = 21
\nAnswer:
\n\"Big<\/p>\n

Question 30.
\nm \u2220 B = 90\u00b0, a = 15, c = 6<\/p>\n

Answer:
\nb = 3\u221a29, \u2220A = 66.9, \u2220C = 23.1<\/p>\n

Explanation:
\nb\u00b2 = a\u00b2 + c\u00b2- 2ac cos B
\nb\u00b2 = 15\u00b2 + 6\u00b2 – 2(15)(6) cos 90
\n= 225 + 36 – 180(0)
\nb\u00b2 = 261
\nb = 3\u221a29
\n\\(\\frac { sin B }{ b } \\) = \\(\\frac { sin A }{ a } \\)
\n\\(\\frac { sin 90 }{ 3\u221a29 } \\) = \\(\\frac { sin A }{ 15 } \\)
\nsin A = 0.92
\n\u2220A = 66.9
\n\u2220A + \u2220B + \u2220C = 180
\n66.9 + 90 + \u2220C = 180
\n\u2220C = 23.1<\/p>\n

Question 31.
\nm \u2220 C = 40\u00b0, b = 27, c = 36
\nAnswer:
\n\"Big<\/p>\n

Question 32.
\na = 34, b = 19, c = 27<\/p>\n

Answer:
\n\u2220B = 33.9, \u2220A = 78.5, \u2220C = 67.6<\/p>\n

Explanation:
\nb\u00b2 = a\u00b2 + c\u00b2- 2ac cos B
\n19\u00b2 = 34\u00b2 + 27\u00b2- 2(34)(27) cos B
\n361 = 1156 + 729 – 1836 cos B
\ncos B = 0.83
\n\u2220B = 33.9
\n\\(\\frac { sin 33.9 }{ 19 } \\) = \\(\\frac { sin A }{ 34 } \\)
\nsin A = 0.98
\n\u2220A = 78.5
\n\u2220A + \u2220B + \u2220C = 180
\n78.5 + 33.9 + \u2220C = 180
\n\u2220C = 67.6<\/p>\n

Question 33.
\nMODELING WITH MATHEMATICS<\/strong>
\nYou and your friend are standing on the baseline of a basketball court. You bounce a basketball to your friend, as shown in the diagram. What is the distance between you and your friend?
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 34.
\nMODELING WITH MATHEMATICS<\/strong>
\nA zip line is constructed across a valley, as shown in the diagram. What is the width w of the valley?
\n\"Big<\/p>\n

Answer:
\nw = 92.5 ft<\/p>\n

Explanation:
\nw\u00b2 = 25\u00b2 + 84\u00b2 – 2(25)(84) cos 102
\nw\u00b2 = 7681 – 4200 cos 102
\nw = 92.5 ft<\/p>\n

Question 35.
\nMODELING WITH MATHEMATICS<\/strong>
\nYou are on the observation deck of the Empire State Building looking at the Chrysler Building. When you turn 145\u00b0 clockwise, you see the Statue of Liberty. You know that the Chrysler Building and the Empire Slate Building arc about 0.6 mile apart and that the Chrysler Building and the Statue of Liberty are about 5.6 miles apart. Estimate the distance between the Empire State Building and the Statue of Liberty.
\nAnswer:
\n\"Big
\n\"Big<\/p>\n

Question 36.
\nMODELING WITH MATHEMATICS<\/strong>
\nThe Leaning Tower of Pisa in Italy has a height of 183 feet and is 4\u00b0 off vertical. Find the horizontal distance d that the top of the tower is off vertical.
\n\"Big<\/p>\n

Answer:<\/p>\n

Question 37.
\nMAKING AN ARGUMENT<\/strong>
\nYour friend says that the Law of Sines can be used to find JK. Your cousin says that the Law of Cosines can be used to find JK. Who is correct\u2019? Explain your reasoning.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 38.
\nREASONING<\/strong>
\nUse \u2206XYZ
\n\"Big
\na. Can you use the Law of Sines to solve \u2206XYZ ? Explain your reasoning.
\nAnswer:<\/p>\n

b. Can you use another method to solve \u2206XYZ ? Explain your reasoning.
\nAnswer:<\/p>\n

Question 39.
\nMAKING AN ARGUMENT<\/strong>
\nYour friend calculates the area of the triangle using the formula A = \\(\\frac{1}{2}\\)qr sin S and says that the area is approximately 208.6 square units. Is your friend correct? Explain your reasoning.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 40.
\nMODELING WITH MATHEMATICS<\/strong>
\nYou are fertilizing a triangular garden. One side of the garden is 62 feet long, and another side is 54 feet long. The angle opposite the 62-foot side is 58\u00b0.
\na. Draw a diagram to represent this situation.
\nb. Use the Law of Sines to solve the triangle from part (a).
\nc. One bag of fertilizer covers an area of 200 square feet. How many bags of fertilizer will you need to cover the entire garden?<\/p>\n

Answer:
\nC = 47.6, A = 74.4, a = 70.4
\n9 bags of fertilizer.<\/p>\n

Question 41.
\nMODELING WITH MATHEMATICS<\/strong>
\nA golfer hits a drive 260 yards on a hole that is 400 yards long. The shot is 15\u00b0 off target.
\n\"Big
\na. What is the distance x from the golfer\u2019s ball to the hole?
\nb. Assume the golfer is able to hit the ball precisely the distance found in part (a). What is the maximum angle \u03b8 (theta) by which the ball can be off target in order to land no more than 10 yards fr0m the hole?
\nAnswer:
\n\"Big
\n\"Big<\/p>\n

Question 42.
\nCOMPARING METHODS<\/strong>
\nA building is constructed on top of a cliff that is 300 meters high. A person standing on level ground below the cliff observes that the angle of elevation to the top of the building is 72\u00b0 and the angle of elevation to the top of the cliff is 63\u00b0.
\na. How far away is the person from the base of the cliff?<\/p>\n

Answer:
\n\"Big<\/p>\n

b. Describe two different methods you can use to find the height of the building. Use one of these methods to find the building\u2019s height.<\/p>\n

Answer:
\nConsider \u25b3SYZ and evaluate d using tangent function
\ntan SYZ = \\(\\frac { 300 }{ d } \\)
\nd = \\(\\frac { 300 }{ tan 63 } \\)
\nd = 152.86
\nThe person is standing 152.86 m away from the base of the cliff.
\nConsider \u25b3XYS and evaluate h + 300
\ntan XYZ = \\(\\frac { h + 300 }{ d } \\)
\nh = 152.86 x tan 72 – 300
\nh = 170.45
\nThe building is 170.45 m high.<\/p>\n

Question 43.
\nMATHEMATICAL CONNECTIONS<\/strong>
\nFind the values of x and y.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 44.
\nHOW DO YOU SEE IT?<\/strong>
\nWould you use the Law of Sines or the Law of Cosines to solve the triangle?
\nAnswer:<\/p>\n

Question 45.
\nREWRITING A FORMULA<\/strong>
\nA Simplify the Law of Cosines for when the given angle is a right angle.
\nAnswer:
\n\"Big<\/p>\n

Question 46.
\nTHOUGHT PROVOKING<\/strong>
\nConsider any triangle with side lengths of a, b, and c. Calculate the value of s, which is half the perimeter of the triangle. What measurement of the triangle is represented by \\(\\sqrt{s(s-a)(s-b)(s-c)} ?\\)
\nAnswer:<\/p>\n

Question 47.
\nANALYZING RELATIONSHIPS<\/strong>
\nThe ambiguous case of the Law of Sines occurs when you are given the measure of one acute angle. the length of one adjacent side, and the length of the side opposite that angle, which is less than the length of the adjacent side. This results in two possible triangles. Using the given information, find two possible solutions for \u2206ABC
\nDraw a diagram for each triangle.
\n(Hint: The inverse sine function gives only acute angle measures. so consider the acute angle and its supplement for \u2220B.)
\n\"Big
\na. m \u2220 A = 40\u00b0, a = 13, b = 16
\nb. m \u2220 A = 21\u00b0, a = 17, b = 32
\nAnswer:
\n\"Big
\n\"Big
\n\"Big<\/p>\n

Question 48.
\nABSTRACT REASONING<\/strong>
\nUse the Law of Cosines to show that the measure of each angle of an equilateral triangle is 60\u00b0. Explain your reasoning.<\/p>\n

Answer:
\na\u00b2 = b\u00b2 + c\u00b2- 2bc cos A
\na\u00b2 = a\u00b2 + a\u00b2 – 2 aa cos A
\na\u00b2 = 2a\u00b2 coas A
\ncos A = 1\/2
\n\u2220A = 60<\/p>\n

Question 49.
\nCRITICAL THINKING<\/strong>
\nAn airplane flies 55\u00b0 east of north from City A to City B. a distance of 470 miles. Another airplane flies 7\u00b0 north of east from City A to City C. a distance of 890 miles. What is the distance between Cities B and C?
\nAnswer:
\n\"Big<\/p>\n

Question 50.
\nREWRITING A FORMULA<\/strong>
\nFollow the steps to derive the formula for the area of a triangle.
\nArea = \\(\\frac{1}{2}\\)ab sin C.
\n\"Big
\na. Draw the altitude from vertex B to \\(\\overline{A C}\\). Label the altitude as h. Write a formula for the area of the triangle using h.
\nAnswer:<\/p>\n

b. Write an equation for sin C
\nAnswer:<\/p>\n

c. Use the results of parts (a) and (b) to write a formula for the area of a triangle that does not include h.
\nAnswer:<\/p>\n

Question 51.
\nPROVING A THEOREM<\/strong>
\nFollow the steps to use the formula for the area of a triangle to prove the Law of Sines (Theorem 9.9).<\/p>\n

a. Use the derivation in Exercise 50 to explain how to derive the three related formulas for the area of a triangle.
\nArea = \\(\\frac{1}{2}\\)bc sin A,
\nArea = \\(\\frac{1}{2}\\)ac sin B,
\nArea = \\(\\frac{1}{2}\\)ab sin C
\nb. why can you use the formulas in part (a) to write the following statement?
\n\\(\\frac{1}{2}\\)bc sin A = \\(\\frac{1}{2}\\)ac sin B = \\(\\frac{1}{2}\\)ab sin C
\nc. Show how to rewrite the statement in part (b) to prove the Law of Sines. Justify each step.
\nAnswer:
\n\"Big
\n\"Big<\/p>\n

Question 52.
\nPROVING A THEOREM<\/strong>
\nUse the given information to complete the two – column proof of the Law of Cosines (Theorem 9.10).
\n\"Big
\nGiven \\(\\overline{B D}\\) is an altitude of \u2206ABC.
\nProve a2<\/sup> = b2<\/sup> + c2<\/sup> – 2bc cos A<\/p>\n\n\n\n\n\n\n\n\n\n\n\n\n
Statements<\/td>\nReasons<\/td>\n<\/tr>\n
1. \\(\\overline{B D}\\) is an altitude of \u2206ABC.<\/td>\n1. Given<\/td>\n<\/tr>\n
2. \u2206ADB and \u2206CDB are right triangles.<\/td>\n2. _______________________<\/td>\n<\/tr>\n
3. a2<\/sup> = (b – x)2<\/sup> + h2<\/sup><\/td>\n3. _______________________<\/td>\n<\/tr>\n
4. _______________________<\/td>\n4. Expand binomial.<\/td>\n<\/tr>\n
5. x2<\/sup> + h2<\/sup> = c2<\/sup><\/td>\n5. _______________________<\/td>\n<\/tr>\n
6. _______________________<\/td>\n6. Substitution Property of Equality<\/td>\n<\/tr>\n
7. cos A = \\(\\frac{x}{c}\\)<\/td>\n7. _______________________<\/td>\n<\/tr>\n
8. x = c cos A<\/td>\n8. _______________________<\/td>\n<\/tr>\n
9. a2<\/sup> = b2<\/sup> + c2<\/sup> – 2bc Cos A<\/td>\n9. _______________________<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Answer:<\/p>\n

Maintaining Mathematical Proficiency<\/p>\n

Find the radius and diameter of the circle.<\/p>\n

Question 53.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 54.
\n\"Big<\/p>\n

Answer:
\nThe radius is 10 in and the diameter is 20 in.<\/p>\n

Question 55.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 56.
\n\"Big<\/p>\n

Answer:
\nThe radius is 50 in and the diameter is 100 in.<\/p>\n

Right Triangles and Trigonometry Review<\/h3>\n

9.1 The Pythagorean Theorem<\/h3>\n

Find the value of x. Then tell whether the side lengths form a Pythagorean triple.<\/p>\n

Question 1.
\n\"Big<\/p>\n

Answer:
\nx = 2\u221a34
\nThe sides will not form a Pythagorean triple.<\/p>\n

Explanation:
\nx\u00b2 = 6\u00b2 + 10\u00b2
\nx\u00b2= 36 + 100
\nx = 2\u221a34<\/p>\n

Question 2.
\n\"Big<\/p>\n

Answer:
\nx = 12
\nThe sides form a Pythagorean triple.<\/p>\n

Explanation:
\n20\u00b2 = 16\u00b2+ x\u00b2
\n400 = 256 + x\u00b2
\nx\u00b2 = 144
\nx = 12<\/p>\n

Question 3.
\n\"Big<\/p>\n

Answer:
\nx = 2\u221a30
\nThe sides will not form a Pythagorean triple.<\/p>\n

Explanation:
\n13\u00b2 = 7\u00b2 + x\u00b2
\n169 = 49 + x\u00b2
\nx = 2\u221a30<\/p>\n

Verify that the segments lengths form a triangle. Is the triangle acute, right, or obtuse?<\/p>\n

Question 4.
\n6, 8, and 9<\/p>\n

Answer:
\n9\u00b2 = 81
\n6\u00b2 + 8\u00b2 = 36 + 64 = 100
\n9\u00b2 < 6\u00b2 + 8\u00b2
\nSo, the triangle is acute<\/p>\n

Question 5.
\n10, 2\u221a2, and 6\u221a3<\/p>\n

Answer:
\n10\u00b2 = 100
\n(2\u221a2)\u00b2 + (6\u221a3)\u00b2 = 8 + 108 = 116
\nSo, the triangle is acute.<\/p>\n

Question 6.
\n13, 18 and 3\u221a55<\/p>\n

Answer:
\n18\u00b2 = 324
\n13\u00b2 + (3\u221a55)\u00b2 = 169 + 495 = 664
\nSo, the triangle is acute.<\/p>\n

9.2 Special Right Triangles<\/h3>\n

Find the value of x. Write your answer in simplest form.<\/p>\n

Question 7.
\n\"Big<\/p>\n

Answer:
\nhypotenuse = leg \u2022 \u221a2
\nx = 6\u221a2<\/p>\n

Question 8.
\n\"Big<\/p>\n

Answer:
\nlonger leg = shorter leg \u2022 \u221a3
\n14 = x \u2022 \u221a3
\nx = 8.08<\/p>\n

Question 9.
\n\"Big<\/p>\n

Answer:
\nlonger leg = shorter leg \u2022 \u221a3
\nx = 8\u221a3 \u2022 \u221a3
\nx = 24<\/p>\n

9.3 Similar Right Triangles<\/h3>\n

Identify the similar triangles. Then find the value of x.<\/p>\n

Question 10.
\n\"Big<\/p>\n

Answer:
\n9 = \u221a(6x)
\n81 = 6x
\nx = \\(\\frac { 27 }{ 2 } \\)<\/p>\n

Question 11.
\n\"Big<\/p>\n

Answer:
\nx = \u221a(6 x 4)
\nx = 2\u221a6<\/p>\n

Question 12.
\n\"Big<\/p>\n

Answer:
\n\\(\\frac { RP }{ RQ } \\) = \\(\\frac { SP }{ RS } \\)
\n\\(\\frac { 9 }{ x } \\) = \\(\\frac { 6 }{ 3 } \\)
\nx = 3.5<\/p>\n

Question 13.
\n\"Big<\/p>\n

Answer:
\n\\(\\frac { SU }{ ST } \\) = \\(\\frac { SV }{ VU } \\)
\n\\(\\frac { 16 }{ 20 } \\) = \\(\\frac { x }{ (x – 16) } \\)
\n4(x – 16) = 5x
\nx = 16<\/p>\n

Find the geometric mean of the two numbers.<\/p>\n

Question 14.
\n9 and 25<\/p>\n

Answer:
\nmean = \u221a(9 x 25)
\n= 15<\/p>\n

Question 15.
\n36 and 48<\/p>\n

Answer:
\nmean = \u221a(36 x 48)
\n= 24\u221a3<\/p>\n

Question 16.
\n12 and 42<\/p>\n

Answer:
\nmean = \u221a(12 x 42)
\n= 6\u221a14<\/p>\n

9.4 The Tangent Ratio<\/h4>\n

Find the tangents of the acute angles in the right triangle. Write each answer as a fraction and as a decimal rounded to four decimal places.<\/p>\n

Question 17.
\n\"Big<\/p>\n

Answer:
\ntan J = \\(\\frac { Opposite side }{ Adjacent side } \\)
\ntan J = \\(\\frac { LK }{ JK } \\) = \\(\\frac { 11 }{ 60 } \\)
\ntan L = \\(\\frac { JK }{ LK } \\) = \\(\\frac { 60 }{ 11 } \\)<\/p>\n

Question 18.
\n\"Big<\/p>\n

Answer:
\ntan P = \\(\\frac { MN }{ MP } \\) = \\(\\frac { 35 }{ 12 } \\)
\ntan N = \\(\\frac { MP }{ MN } \\) = \\(\\frac { 12 }{ 35 } \\)<\/p>\n

Question 19.
\n\"Big<\/p>\n

Answer:
\ntan A = \\(\\frac { BC }{ AC } \\) = \\(\\frac { 7 }{ 4\u221a2 } \\)
\ntan B = \\(\\frac { AC }{ BC } \\) = \\(\\frac { 4\u221a2 }{ 7 } \\)<\/p>\n

Find the value of x. Round your answer to the nearest tenth.<\/p>\n

Question 20.
\n\"Big<\/p>\n

Answer:
\ntan 54 = \\(\\frac { x }{ 32 } \\)
\n1.37 = \\(\\frac { x }{ 32 } \\)
\nx = 43.8<\/p>\n

Question 21.
\n\"Big<\/p>\n

Answer:
\ntan 25 = \\(\\frac { x }{ 20 } \\)
\n0.46 x 20 = x
\nx = 9.2<\/p>\n

Question 22.
\n\"Big<\/p>\n

Answer:
\ntan 38 = \\(\\frac { 10 }{ x } \\)
\nx = 12.82<\/p>\n

Question 23.
\nThe angle between the bottom of a fence and the top of a tree is 75\u00b0. The tree is 4 let from the fence. How tall is the tree? Round your answer to the nearest foot.
\n\"Big<\/p>\n

Answer:
\ntan 75 = \\(\\frac { x }{ 4 } \\)
\nx = 14.92<\/p>\n

9.5 The Sine and Cosine Ratios<\/h3>\n

Find sin X, sin Z, cos X, and cos Z. Write each answer as a fraction and as a decimal rounded to four decimal places.<\/p>\n

Question 24.
\n\"Big<\/p>\n

Answer:
\nsin X = \\(\\frac { 3 }{ 5 } \\)
\nsin Z = \\(\\frac { 4 }{ 5 } \\)
\ncos X = \\(\\frac { 4 }{ 5 } \\)
\ncos Z = \\(\\frac { 3 }{ 5 } \\)<\/p>\n

Question 25.
\n\"Big<\/p>\n

Answer:
\nsin X = \\(\\frac { 7 }{ \u221a149 } \\)
\nsin Z = \\(\\frac { 10 }{ \u221a149 } \\)
\ncos X = \\(\\frac { 10 }{ \u221a149 } \\)
\ncos Z = \\(\\frac { 7 }{ \u221a149 } \\)<\/p>\n

Question 26.
\n\"Big<\/p>\n

Answer:
\nsin X = \\(\\frac { 55 }{ 73 } \\)
\nsin Z = \\(\\frac { 48 }{ 73 } \\)
\ncos X = \\(\\frac { 48 }{ 73 } \\)
\ncos Z = \\(\\frac { 55 }{ 73 } \\)<\/p>\n

Find the value of each variable using sine and cosine. Round your answers to the nearest tenth.<\/p>\n

Question 27.
\n\"Big<\/p>\n

Answer:
\nsin 23 = \\(\\frac { t }{ 34 } \\)
\nt = 13.26
\ncos 23 = \\(\\frac { s }{ 34 } \\)
\ns = 31.28<\/p>\n

Question 28.
\n\"Big<\/p>\n

Answer:
\nsin 36 = \\(\\frac { s }{ 5 } \\)
\ns = 2.9
\ncos 36 = \\(\\frac { r }{ 5 } \\)
\nr = 4<\/p>\n

Question 29.
\n\"Big<\/p>\n

Answer:
\nsin 70 = \\(\\frac { v }{ 10 } \\)
\nv = 9.39
\ncos 70 = \\(\\frac { w }{ 10 } \\)
\nw = 3.42<\/p>\n

Question 30.
\nWrite sin 72\u00b0 in terms of cosine.<\/p>\n

Answer:
\nsin 72 = cos(90 – 72)
\n= cos 18 = 0.95<\/p>\n

Question 31.
\nWrite cos 29\u00b0 in terms of sine.<\/p>\n

Answer:
\nsin 29 = cos(90 – 29)
\n= cos 61 = 0.48<\/p>\n

9.6 Solving Right Triangles<\/h3>\n

Let \u2220Q be an acute angle. Use a calculator to approximate the measure of \u2220Q to the nearest tenth of a degree.<\/p>\n

Question 32.
\ncos Q = 0.32<\/p>\n

Answer:
\ncos Q = 0.32
\n\u2220Q = inverse cos of .32
\n\u2220Q = 71.3<\/p>\n

Question 33.
\nsin Q = 0.91<\/p>\n

Answer:
\nsin Q = 0.91
\n\u2220Q = inverse sin of 0.91
\n\u2220Q = 65.5<\/p>\n

Question 34.
\ntan Q = 0.04<\/p>\n

Answer:
\ntan Q = 0.04
\n\u2220Q = inverse tan of 0.04
\n\u2220Q = 2.29<\/p>\n

Solve the right triangle. Round decimal answers to the nearest tenth.<\/p>\n

Question 35.
\n\"Big<\/p>\n

Answer:
\na = 5\u221a5, \u2220A = 47.7, \u2220B = 42.3<\/p>\n

Explanation:
\nc\u00b2 = a\u00b2 + b\u00b2
\n15\u00b2 = a\u00b2 + 10\u00b2
\na\u00b2 = 125
\na = 5\u221a5
\nsin A = \\(\\frac { 5\u221a5 }{ 15 } \\) = 0.74
\n\u2220A = 47.7
\n\u2220A + \u2220B + \u2220C = 180
\n47.7 + \u2220B + 90 = 180
\n\u2220B = 42.3<\/p>\n

Question 36.
\n\"Big<\/p>\n

Answer:
\nNL = 7.59, \u2220L = 53, ML = 4.55<\/p>\n

Explanation:
\ncos 37 = \\(\\frac { 6 }{ NL } \\)
\nNL = 7.59
\n\u2220N + \u2220M + \u2220L = 180
\n37 + 90 + \u2220L = 180
\n\u2220L = 53
\nsin 37 = \\(\\frac { ML }{ 7.59 } \\)
\nML = 4.55<\/p>\n

Question 37.
\n\"Big<\/p>\n

Answer:
\nXY = 17.34, \u2220X = 46, \u2220Z = 44<\/p>\n

Explanation:
\nc\u00b2 = a\u00b2 + b\u00b2
\n25\u00b2 = 18\u00b2+ b\u00b2
\nb\u00b2 = 301
\nb = 17.34
\nsin X = \\(\\frac { 18 }{ 25 } \\)
\n\u2220X = 46
\nsum of angles = 180
\n46 + 90 + \u2220Z = 180
\n\u2220Z = 44<\/p>\n

9.7 Law of Sines and Law of Cosines<\/h3>\n

Find the area of \u2206ABC with the given side lengths and included angle.<\/p>\n

Question 38.
\nm \u2220 B = 124\u00b0, a = 9, c = 11<\/p>\n

Answer:
\nArea = \\(\\frac { 1 }{ 2 } \\) ac sin B
\n= \\(\\frac { 1 }{ 2 } \\) (9 x 11) sin 124
\n= 40.59<\/p>\n

Question 39.
\nm \u2220 A = 68\u00b0, b = 13, c = 7<\/p>\n

Answer:
\nArea = \\(\\frac { 1 }{ 2 } \\) bc sin A
\n= \\(\\frac { 1 }{ 2 } \\) (13 x 7) sin 68
\n= 41.86<\/p>\n

Question 40.
\nm \u2220 C = 79\u00b0, a = 25 b = 17<\/p>\n

Answer:
\nArea = \\(\\frac { 1 }{ 2 } \\) ab sin C
\n= \\(\\frac { 1 }{ 2 } \\) (25 x 17) sin 79
\n= 208.25<\/p>\n

Solve \u2206ABC. Round decimal answers to the nearest tenth.<\/p>\n

Question 41.
\nm \u2220 A = 112\u00b0, a = 9, b = 4<\/p>\n

Answer:
\n\u2220B = 24, \u2220C = 44, c = 6.76<\/p>\n

Explanation:
\n\\(\\frac { sin B }{ b } \\) = \\(\\frac { sin A }{ a } \\)
\n\\(\\frac { sin B }{ 4 } \\) = \\(\\frac { sin 112 }{ 9 } \\)
\nsin B = 0.408
\n\u2220B = 24
\n\u2220A + \u2220B + \u2220C = 180
\n112 + 24 + \u2220C = 180
\n\u2220C = 44
\n\\(\\frac { sin 112 }{ 9 } \\) = \\(\\frac { sin 44 }{ c } \\)
\nc = 6.76<\/p>\n

Question 42.
\nm \u2220 4 = 28\u00b0, m \u2220 B = 64\u00b0, c = 55<\/p>\n

Answer:
\n\u2220C = 88, b = 49.4, a = 25.5<\/p>\n

Explanation:
\n\u2220A + \u2220B + \u2220C = 180
\n28 + 64 + \u2220C = 180
\n\u2220C = 88
\n\\(\\frac { sin B }{ b } \\) = \\(\\frac { sin C }{ c } \\)
\n\\(\\frac { sin 64 }{ b } \\) = \\(\\frac { sin 88 }{ 55 } \\)
\n\\(\\frac { sin 64 }{ b } \\) = 0.018
\nb = 49.4
\n\\(\\frac { sin 28 }{ a } \\) = \\(\\frac { sin 88 }{ 55 } \\)
\na = 25.5<\/p>\n

Question 43.
\nm \u2220 C = 48\u00b0, b = 20, c = 28<\/p>\n

Answer:
\n\u2220B = 31.3, \u2220A = 100.7, a = 37.6<\/p>\n

Explanation:
\n\\(\\frac { sin B }{ b } \\) = \\(\\frac { sin C }{ c } \\)
\n\\(\\frac { sin B }{ 20 } \\) = \\(\\frac { sin 48 }{ 28 } \\)
\n\\(\\frac { sin B }{ 20 } \\) = 0.026
\nsin B = 0.52
\n\u2220B = 31.3
\n\u2220A + \u2220B + \u2220C = 180
\n\u2220A + 31.3 + 48 = 180
\n\u2220A = 100.7
\n\\(\\frac { sin 100.7 }{ a } \\) = \\(\\frac { sin 48 }{ 28 } \\)
\na = 37.6<\/p>\n

Question 44.
\nm \u2220 B = 25\u00b0, a = 8, c = 3<\/p>\n

Answer:
\nb = 5.45, \u2220A = 37.5, \u2220C = 117.5<\/p>\n

Explanation:
\nb\u00b2 = a\u00b2 + c\u00b2- 2ac cos B
\nb\u00b2 = 8\u00b2 + 3\u00b2 – 2(8 x 3) cos 25 = 73 – 43.2 = 29.8
\nb = 5.45
\n\\(\\frac { sin A }{ 8 } \\) = \\(\\frac { sin 25 }{ 5.45 } \\)
\nsin A = 0.61
\n\u2220A = 37.5
\n\u2220A + \u2220B + \u2220C = 180
\n37.5 + 25 + \u2220C = 180
\n\u2220C = 117.5<\/p>\n

Question 45.
\nm \u2220 B = 102\u00b0, m \u2220 C = 43\u00b0, b = 21<\/p>\n

Answer:
\n\u2220A = 35, c = 14.72, a = 12.3<\/p>\n

Explanation:
\n\u2220A + \u2220B + \u2220C = 180
\n\u2220A + 102 + 43 = 180
\n\u2220A = 35
\n\\(\\frac { sin 102 }{ 21 } \\) = \\(\\frac { sin 43 }{ c } \\)
\n0.046 = \\(\\frac { sin 43 }{ c } \\)
\nc = 14.72
\n\\(\\frac { sin 102 }{ 21 } \\) = \\(\\frac { sin 35 }{ a } \\)
\n0.046 = \\(\\frac { sin 35 }{ a } \\)
\na = 12.3<\/p>\n

Question 46.
\na = 10, b = 3, c = 12<\/p>\n

Answer:
\n\u2220B = 11.7, \u2220C = 125.19, \u2220A = 43.11<\/p>\n

Explanation:
\nb\u00b2 = a\u00b2 + c\u00b2- 2ac cos B
\n3\u00b2 = 10\u00b2 + 12\u00b2- 2(10 x 12) cos B
\n9 = 100 + 144 – 240 cos B
\ncos B = 0.979
\n\u2220B = 11.7
\na\u00b2 = b\u00b2 + c\u00b2- 2bc cos A
\n100 = 9 + 144 – 72 cos A
\ncos A = 0.73
\n\u2220A = 43.11
\n\u2220A + \u2220B + \u2220C = 180
\n43.11 + 11.7 + \u2220C = 180
\n\u2220C = 125.19<\/p>\n

Right Triangles and Trigonometry Test<\/h3>\n

Find the value of each variable. Round your answers to the nearest tenth.<\/p>\n

Question 1.
\n\"Big
\nAnswer:
\nsin 25 = \\(\\frac { t }{ 18 } \\)
\nt = 7.5
\ncos 25 = \\(\\frac { s }{ 18 } \\)
\ns = 16.2<\/p>\n

Question 2.
\n\"Big
\nAnswer:
\nsin 22 = \\(\\frac { 6 }{ x } \\)
\nx = 16.21
\ncos 22 = \\(\\frac { y }{ 16.21 } \\)
\ny = 14.91<\/p>\n

Question 3.
\n\"Big
\nAnswer:
\ntan 40 = \\(\\frac { k }{ 10 } \\)
\nk = 8.3
\ncos 40 = \\(\\frac { 10 }{ j } \\)
\nj = 13.15<\/p>\n

Verity that the segment lengths form a triangle. Is the triangle acute, right, or obtuse?<\/p>\n

Question 4.
\n16, 30, and 34<\/p>\n

Answer:
\n34\u00b2= 16\u00b2 + 30\u00b2
\nSo, the triangle is a right-angled triangle.<\/p>\n

Question 5.
\n4, \u221a67, and 9<\/p>\n

Answer:
\n9\u00b2 = 81
\n4\u00b2 + (\u221a67)\u00b2 = 83
\nSo the triangle is acute<\/p>\n

Question 6.
\n\u221a5. 5. and 5.5<\/p>\n

Answer:
\n5.5\u00b2 = 30.25
\n\u221a5\u00b2 + 5\u00b2 = 30
\nSo the triangle is obtuse<\/p>\n

Solve \u2206ABC. Round decimal answers to the nearest tenth.<\/p>\n

Question 7.
\n\"Big
\nAnswer:
\nc = 12.08, \u2220A = 24.22, \u2220C = 65.78<\/p>\n

Explanation:
\ntan A = \\(\\frac { 5 }{ 11 } \\)
\n\u2220A = 24.22
\nc\u00b2 = 11\u00b2+ 5\u00b2
\nc = 12.08
\n24.22 + 90 + \u2220C = 180
\n\u2220C = 65.78<\/p>\n

Question 8.
\n\"Big<\/p>\n

Answer:
\n\u2220B = 35.4, \u2220C = 71.6, c = 17.9<\/p>\n

Explanation:
\n\\(\\frac { sin 73 }{ 18 } \\) = \\(\\frac { sin B }{ 11 } \\)
\nsin B = 0.58
\n\u2220B = 35.4
\n73 + 35.4 + \u2220C = 180
\n\u2220C = 71.6
\n\\(\\frac { sin 73 }{ 18 } \\) = \\(\\frac { sin 71.6 }{ c } \\)
\nc = 17.9<\/p>\n

Question 9.
\n\"Big<\/p>\n

Answer:
\nBC = 4.54, \u2220B = 59.3, \u2220A = 30.7<\/p>\n

Explanation:
\n9.2\u00b2 = 8\u00b2 + x\u00b2
\nx = 4.54
\n\\(\\frac { sin 90 }{ 9.2 } \\) = \\(\\frac { sin B }{ 8 } \\)
\nsin B = 0.86
\n\u2220B = 59.3
\n\u2220A + 59.3 + 90 = 180
\n\u2220A = 30.7<\/p>\n

Question 10.
\nm \u2220 A = 103\u00b0, b = 12, c = 24<\/p>\n

Answer:
\na = 29, \u2220B = 53.5<\/p>\n

Explanation:
\na\u00b2 = b\u00b2 + c\u00b2- 2bc cos A
\na\u00b2 = 144 + 24\u00b2 – 2(12 x 24) cos 103
\na = 29
\n\\(\\frac { sin B }{ 12 } \\) = \\(\\frac { sin 103 }{ 29 } \\)
\n\u2220B = 23.5
\n\u2220C = 180 – (103 + 23.5) = 53.5<\/p>\n

Question 11.
\nm \u2220 A = 26\u00b0, m \u2220 C = 35\u00b0, b = 13<\/p>\n

Answer:
\n\u2220B = 119, a = 6.42, c = 8.5<\/p>\n

Explanation:
\n\u2220B + 26 + 35 = 180
\n\u2220B = 119
\n\\(\\frac { sin 119 }{ 13 } \\) = \\(\\frac { sin 26 }{ a } \\)
\na = 6.42
\n\\(\\frac { sin 119 }{ 13 } \\) = \\(\\frac { sin 35 }{ c } \\)
\nc = 8.5<\/p>\n

Question 12.
\na = 38, b = 31, c = 35<\/p>\n

Answer:
\n\u2220B=50.2, \u2220C = 59.8, \u2220A = 70<\/p>\n

Explanation:
\nb\u00b2 = a\u00b2 + c\u00b2- 2ac cos B
\n31\u00b2 = 38\u00b2+ 35\u00b2- 2(35 x 38) cos B
\ncos B = 0.64
\n\u2220B=50.2
\na\u00b2 = b\u00b2 + c\u00b2- 2bc cos A
\n38\u00b2 = 31\u00b2+ 35\u00b2- 2(31 x 35) cos A
\ncos A = 0.341
\n\u2220A = 70
\n\u2220C = 59.8<\/p>\n

Question 13.
\nWrite cos 53\u00b0 in terms of sine.<\/p>\n

Answer:
\ncos 53\u00b0 = sin (90 – 53) = sin 37<\/p>\n

Find the value of each variable. Write your answers in simplest form.<\/p>\n

Question 14.
\n\"Big
\nAnswer:
\nsin 45 = \\(\\frac { 16 }{ q } \\)
\nq = 22.6
\ncos 45 = \\(\\frac { r }{ q } \\)
\nr = 16<\/p>\n

Question 15.
\n\"Big
\nAnswer:<\/p>\n

Question 16.
\n\"Big
\nAnswer:
\nsin 30 = \\(\\frac { f }{ 9.2 } \\)
\nf = 4.6
\ncos 30 = \\(\\frac { 8 }{ h } \\)
\nh = 9.2<\/p>\n

Question 17.
\nIn \u2206QRS, m \u2220 R = 57\u00b0, q = 9, and s = 5. Find the area of \u2206QRS.<\/p>\n

Answer:
\nArea = \\(\\frac { 1 }{ 2 } \\) qs sin R
\n= \\(\\frac { 1 }{ 2 } \\) (9 x 5) sin 57 = 18.675<\/p>\n

Question 18.
\nYou are given the measures of both acute angles of a right triangle. Can you determine the side lengths? Explain.<\/p>\n

Answer:
\nNo.<\/p>\n

Question 19.
\nYou are at a parade looking up at a large balloon floating directly above the street. You are 60 feet from a point on the street directly beneath the balloon. To see the top of the balloon, you look up at an angle of 53\u00b0. To see the bottom of the balloon, you look up at an angle of 29\u00b0. Estimate the height h of the balloon.
\n\"Big<\/p>\n

Answer:<\/p>\n

Question 20.
\nYou warn to take a picture of a statue on Easter Island, called a moai. The moai is about 13 feet tall. Your camera is on a tripod that is 5 feet tall. The vertical viewing angle of your camera is set at 90\u00b0. How far from the moai should you stand so that the entire height of the moai is perfectly framed in the photo?
\n\"Big<\/p>\n

Answer:<\/p>\n

Right Triangles and Trigonometry Cummulative Assessment<\/h3>\n

Question 1.
\nThe size of a laptop screen is measured by the length of its diagonal. you Want to purchase a laptop with the largest screen possible. Which laptop should you buy?
\n(A)
\n\"Big
\n(B)
\n\"Big
\n(C)
\n\"Big
\n(D)
\n\"Big
\nAnswer:
\n(B)<\/p>\n

Explanation:
\n(a) d = \u221a9\u00b2 + 12\u00b2 = 15
\n(b) d = \u221a11.25\u00b2 + 20\u00b2 = 22.94
\n(c) d = \u221a12\u00b2 + 6.75\u00b2 = 13.76
\n(d) d = \u221a8\u00b2 + 6\u00b2 = 10<\/p>\n

Question 2.
\nIn \u2206PQR and \u2206SQT, S is between P and Q, T is between R and Q, and \\(\\) What must be true about \\(\\overline{S T}\\) and \\(\\overline{P R}\\)? Select all that apply.
\n\\(\\overline{S T}\\) \u22a5 \\(\\overline{P R}\\)\u00a0 \u00a0 \u00a0 \\(\\overline{S T}\\) || \\(\\overline{P R}\\)\u00a0 \u00a0 ST = PR\u00a0 \u00a0 \u00a0 \u00a0 ST = \\(\\frac{1}{2}\\)PR
\nAnswer:<\/p>\n

Question 3.
\nIn the diagram, \u2206JKL ~ \u2206QRS. Choose the symbol that makes each statement true.
\n\"Big
\n<\u00a0 \u00a0 \u00a0 =\u00a0 \u00a0 \u00a0 \u00a0>
\nsin J ___________ sin Q\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0sin L ___________ cos J \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 cos L ___________ tan Q
\ncos S ___________ cos\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 J cos J ___________ sin S\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0tan J ___________ tan Q
\ntan L ___________ tan Q\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0tan S ___________ cos Q\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 sin Q ___________ cos L
\nAnswer:
\nsin J = sin Q\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0sin L = cos J \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 cos L = tan Q
\ncos S > cos J\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 cos J > sin S\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0tan J = tan Q
\ntan L < tan Q\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0tan S > cos Q\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 sin Q = cos L<\/p>\n

Question 4.
\nA surveyor makes the measurements shown. What is the width of the river.
\n\"Big<\/p>\n

Answer:
\ntan 34 = \\(\\frac { AB }{ 84 } \\)
\nAB = 56.28<\/p>\n

Question 5.
\nCreate as many true equations as possible.
\n\"Big<\/p>\n

____________ = ______________<\/p>\n

sin X\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 cos X \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 tan x \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \\(\\frac{X Y}{X Z}\\) \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \\(\\frac{Y Z}{X Z}\\)<\/p>\n

Sin Z \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0cos Z \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 tan Z \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \\(\\frac{X Y}{Y Z}\\) \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \\(\\frac{Y Z}{X Y}\\)<\/p>\n

Answer:
\nsin X = \\(\\frac{Y Z}{X Z}\\) = cos Z
\ncos X = \\(\\frac{X Y}{X Z}\\) = sin Z
\ntan x = \\(\\frac{Y Z}{X Y}\\)
\ntan Z = \\(\\frac{X Y}{Y Z}\\)<\/p>\n

Question 6.
\nProve that quadrilateral DEFG is a kite.
\nGiven \\(\\overline{H E} \\cong \\overline{H G}\\), \\(\\overline{E G}\\) \u22a5 \\(\\overline{D F}\\)
\nProve \\(\\overline{F E} \\cong \\overline{F G}\\), \\(\\overline{D E} \\cong \\overline{D G}\\)
\n\"Big
\nAnswer:<\/p>\n

Question 7.
\nWhat are the coordinates of the vertices of the image of \u2206QRS after the composition of transformations show?
\n\"Big
\n(A) Q’ (1, 2), R'(5, 4), S'(4, -1)
\n(B) Q'(- 1, – 2), R’ (- 5, – 4), S’ (- 4, 1)
\n(C) Q'(3, – 2), R’ (- 1, – 4), S’ (0, 1)
\n(D) Q’ (-2, 1), R'(- 4, 5), S'(1, 4)
\nAnswer:<\/p>\n

Question 8.
\nThe Red Pyramid in Egypt has a square base. Each side of the base measures 722 feet. The height of the pyramid is 343 fee.
\n\"Big<\/p>\n

a. Use the side length of the base, the height of the pyramid, and the Pythagorean Theorem to find the slant height, AB, of the pyramid.<\/p>\n

Answer:
\n343\u00b2 = h\u00b2 + 722\u00b2
\nh = 635.3<\/p>\n

b. Find AC.
\nAnswer:
\nAC = 343<\/p>\n

c. Name three possible ways of finding m \u2220 1. Then, find m \u2220 1.
\nAnswer:
\nThree possible ways are sin 1, cos 1 and tan 1
\ntan 1 = \\(\\frac { 722 }{ 635.3 } \\)
\n\u22201 = 48<\/p>\n","protected":false},"excerpt":{"rendered":"

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