Common Core Math

Discount is defined as the amount or percentage that deducted or reduced from the normal selling price of any product. Shortly, discount means the reduction in the price of a good or service. Refer to the marked price, list price and discount, and the formula of discount, and procedure to find out the discount, and solved examples in this article and get a grip on the total discount concept. Subtracting the Sale price from the Regular price minus gives the amount of discount. Check out various problems on discount and improve your math knowledge easily.

Discount Formula

Check out the formula of discount and Rate of Discount in the below section. Use the below formulas to find different problems on discount.

• Discount = List Price – Selling Price
• Therefore, Selling Price = List Price + Discount
• List Price = Selling Price + Discount

Discount Percentage Formula is

• Rate of Discount or Discount% = (Discount/List Price) * 100
• Selling Price = List Price ((100 – Discount)/100)%
• List Price = Selling Price (100/(100-Discount)%)

Concept of Discount

Sometimes, the retailers don’t directly sell the old items, defective items. They offer less price for those products to sell. Such a price is called the sale price. The difference between the Marked price (selling price) and the sale price is called a discount.
Discount = Marked Price – Sale Price

Marked Price

The price on the label of a product is called the Marked Price. The marked price is the price at which the product ready to be sold. Marked price is represented by MP.
Example: The price printed on the books is called the Marked Price.

List Price

The products that are manufactured in a factory and marked with a price by the retailer to sell them are called the List Price. The list price is represented as LP.

Selling price = Marked price – Discount
where the selling price is the amount that actually pays for the product when you purchase.
The marked price is the general price of the product without any discount.
Discount is a percentage of the Marked price.

Example:
If the cost of the pen is $10 and the shopkeeper reducing the amount of $8 on that pen, then there is a discount of $2 is available for that pen.
Sometimes, the discount is also available in percentages ‘%’.

Do Check Related Articles:

• Practice Test on Profit Loss and Discount
• Concept of Profit and Loss
• Calculate Profit and Profit Percent

How to Find Discount?

The below procedure will help you to find a discount on a product. Follow the procedure and practice different problems on discount using the below process.

1. The rate given for a product is considered as a percentage value.
2. In the next step, to find out the discount, multiply the rate by the original price.
3. To find the sale price, Subtract the discount from the original price.

Types of Discounts

Basically, Discounts are classified into three types. They are

Trade Discounts: The trade discounts are offered by the distributors. Trade discounts are the discounts where the amount of the product is reduced to sell the product.
Quantity Discounts: Quantity discounts are applied to a buyer depending on the number of items he purchased in greater numbers. Mainly, to sell the items in huge amounts to buyers, the sellers offer Quantity discounts.
Promotional Discounts: Promotional discounts are applied to do sales promotions. It is most preferable amongst shopkeepers. Generally, the Promotional discounts appear in % or buy 2 & get 1 free, etc.

Discount Examples

1. The marked price of a table fan is $ 750 and the shopkeeper offers a discount of 5% on it. Find the selling price of the table fan?

Solution:
Given that the marked price of a table fan is $ 750 and the shopkeeper offers a discount of 5% on it.
The Marked Price of the table fan = $750.
Discount on the table fan = 5%
We know that Discount = 5% of Marked Price
Discount = 5% of $ 750 = 750 * \(\frac { 5 }{ 100 } \) = 750 * 0.05 = 37.5
Discount = $37.5
To find the selling price of the table fan, subtract the discount from the Marked Price of the table fan.
Selling price = (Marked Price) – (discount)
Selling price = $750 – $37.5 = $712.5

Hence, the selling price of the table fan is $712.5.

2. A trader fixed the cost of his goods at 30% above the cost price and allows a discount of 15%. What is his gain percent?

Solution:
Given that a trader fixed the cost of his goods at 30% above the cost price and allows a discount of 15%.
Let the cost price be $ 100.
Therefore, the marked price = $130
We know that Discount = 15% of Marked Price
Discount = 15% of $130 = 130 * \(\frac { 15 }{ 100 } \) = 19.5
Discount =$19.5
To find the selling price of the table fan, subtract the discount from the Marked Price of the table fan.
Selling price = (Marked Price) – (discount)
Selling price = $130 – $19.5 = $110.5
Gain% = (Selling price – cost price)% = (110.5 – 100)% = 10.5%

Hence, the trader gains 10.5%.

3. A dealer purchased a cooler for $ 3830. He allows a discount of 6% on its marked price and still gains 5%. Find the marked price of the cooler?

Solution:
Given that a dealer purchased a cooler for $ 3830. He allows a discount of 6% on its marked price and still gains 5%.
The Cost price of the cooler = $ 3830
Gain% = 5%.
Now, find out the selling price.
To find the selling price, Substitute the Cost price and Gain% in the selling price formula.
Therefore, selling price = [{(100 + gain%)/100} × CP]
Selling price = $ [{(100 + 5)/100} × 3830] = $4021.5
The selling price is $4021.5
Let the marked price be $ x.
Then, the discount = 6% of marked price = 6% of $ x.
discount = $ {x × (6/100)} = = $ 3x/50
To find the selling price of the table fan, subtract the discount from the Marked Price of the table fan.
Selling price = (Marked Price) – (discount)
Selling price = $ x – $ 3x/50 = $47x/50
But, the SP = $4021.5
Therefore, $47x/50 = $4021.5
$47x = $4021.5 * 50
$47x = 201075
x = 201075/47 = $4278.19

Hence, the Marked price of the cooler is $4278.19.

4. How much percent above the cost price should a shopkeeper mark his goods so that after allowing a discount of 15% on the marked price, he gains 30%?

Solution:
Let the cost price be $ 100.
Gain required = 30%.
Therefore, the selling price = $ 130.
Let the marked price be $x.
Then, discount = 15% of $x
discount = $ (x × 15/100)
discount = $ 3x/20
To find the selling price of the table fan, subtract the discount from the Marked Price of the table fan.
Therefore, selling price = (Marked Price) – (discount)
= $ {x – ($ 3x/20)
= $ 17x/20
Therefore, $ 17x/20 = 130
⇔ x = {130 × (20/17)} = 152.94
Therefore, marked price = $ 152.94.

Hence, the marked price is 52.94% above the cost price.

Successive Discounts

If two or more discounts are applied one after the other on a single product, then such discounts are called Successive Discounts. The Successive Discounts are also called discounts in series.

For example, if a discount of 30% is given for a product and later 10% of discount is given due to the reduced price. In such a situation, we can say successive discounts of 30% and 10% are given.

5. Find the single discount equivalent to two successive discounts of 30% and 20%.

Solution:
Let the marked price of the product be $ 100.
Then, the first discount on it = $ 30.
Price after first discount = $ (100 – 30) = $ 70.
Second discount on it = 20% of $ 70
= $ {70 × (20/100)} = $ 14.
Price after second discount = $ (70 – 14) = $ 56.
Net selling price = $ 56.

Single discount equivalent to given successive discounts = (100 – 56)% = 44%.

Fake Discounts

Fake Discounts also exist where the product pre-sale price may immoderately increase, or else the post-sale price of the product is actually its market price. This may feel the customers that they are getting a discount to make them purchase a product.

Frequently Asked Questions on Discounts

1. What is the Discount?

Discount is the amount deducted from the list price of the commodity before selling it to the customer.

2. How to Calculate Discount?

Check out the below steps to calculate the discount.
1. Find out the original price of the product.
2. Know the discount percentage.
3. Calculate the savings.
4. Calculate the sale price by subtracting the savings from the original price.
To calculate the discount, we need to subtract the selling price from the list price.
Discount = Selling Price – List Price

3. What is the Discounted Rate?

The discounted rate is also called the discount percentage. It is defined as the percentage by which list price is reduced before selling.

4. What is the Discount Percentage Formula?

The formula for Discount Percentage is
Rate of Discount or Discount% = (Discount/List Price) * 100
Selling Price = List Price ((100 – Discount)/100)%
List Price = Selling Price (100/(100-Discount)%)

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Are you a beginner to learn the concept of Skip Counting by 3S? If yes, then check out here to know each and every detail about this concept. Skip counting is considered a skill and also viewed as an arbitrary math skill. With the help of skip counting facts and definitions, you can improve learning the multiplication facts and along with the number sense. Skip counting deals with counting the numbers by 2S, 3S, 5S, etc Here in the below sections, we will see a detailed explanation on Skip Counting by 3S. Follow the facts, definitions, examples, etc.

Skip Counting by 3S – Introduction

Before, going to know how to skip count, first, we will learn about what is skip counting. Skip Counting is a technique or method of counting numbers other than 1. In the method of skip counting, we add the same numbers to the previous number each time.

Example:

Find the series of numbers with the skip count by 2?

Solution:

After applying the skip count by 2

we get the number series as 2,4,6,8,10,12…

Therefore, we get a series of even numbers.

Important Points of Skip Counting

• Skip Counting on any number is possible.
• This helps the numbers to count easily.
• The skip counting method involves a huge application in the multiplication of tables.
• Skip counting can be done with any numbers like 2S, 3S, 5S, 10S, 100S, etc.

Skip Counting by 3S – Important Points

• To skip count by number 3, keep adding the value 3 to get the next number.
• We can notice the diagonal patterns in the number grid when applying the skip counting by 3.
• Use the grid for the skip count and we miss out on two numbers and write the third digit.
• Skip counting by 3S from zero is essential in learning the 3’s timetable.
• For skip counting by 3S, we get the number line as 3,6,9,15,18,21,24,27,30 etc
• Learning the counting by 3 patterns till to 30 will help you with counting in 3’s with the larger numbers.

How to Skip Counting by 3?

Skip counting is the fastest way to count than simply counting one by one.

Skip counting by 3S means, adding the three numbers to get the next number, which means that we skip two numbers out in the process.

The number series of skip counting by 3S from zero, we have:

0,3,6,9,12,15,18,21,24,27,30

Since the pattern is not clear, the number grid will help you to understand the counting by three easily. With the help of the number grid, we can easily see the numbers that come next, and also we can see the numbers we skip while forming the number series.

To notice the pattern in the 3’s timetable, we have to look at the units digit column for the numbers.

Consider 30, we check for the units digit of 30 i.e., 0

Therefore, 0+3=3, hence, 30 + 3 = 33

Continuing to add three numbers, we get

30,33,36,39,42,45,48,51,54,57,60

Forward Skip Counting

In the forward skip counting, we count the numbers in the forward direction. It means that we skip the count of numbers for positive values. Counting the numbers with skip values has major applications in the real life. For suppose, if we want to count 100 marbles, use the skip counting method. If we count one-one marble, it consumes a lot of time. Thus, we use the skip count method, by any of the big numbers like 10 or 20, then we will count them quickly.

Backward Skip Counting

Backward skip counting is also essential for the students which implies counting to the negative numbers. For example, if we want to count by -2 then the number series is

-2,-4,-6,-8,-10 etc.

Strategies to Skip Counting

1. Questioning

A much-overlooked strategy of teaching is questioning. There are various questions that can give you clarity on the topic. To skip count by 2S, try asking the question like “In how many ways you can count to 24?” or you can also ask to “Can you count to 40 by 10’s? Why or Whynot?”

2. Use a Calculator

Generally, we don’t use calculators when solving skip counting problems. However, it can be a useful and fun tool when solving this skill. First of all, practice how to change the numbers by pressing the equal sign. This kind of strategy helps to build the number fluency and also to remember the skip counting values.

3. Play Game

This kind of strategy is used for kids. This is an easy process where kids can remember skip counting and also can have fun. Make the students stand or sit in a circle. Stand counting the student with a random number and then they continue to count by skipping the number until they go the complete way around the circle.

4. Use Manipulatives

To remember the method and rules easily, make use of manipulatives like paper clips, candy, snap cubes. With the help of these manipulatives, you can recite a string of numbers and can understand the purpose of skip counting.

Skip Counting by 2S

In this method, we add the number 2 for each counting. In this manner, we get alternate numbers. For example, if we start counting from 2, then by skipping the count of numbers, we get the number series as

2,4,6,8,10,12,14,16,18,20,22…

Skip Counting by 4

When we skip count by 4, then we have to add 4 in each step. The procedure we follow for skip count 4 is

0+4 = 4, 4+4 = 8, 8+4 = 12, 12+4=16, 16+4 = 20, 20+4 = 24,24+4 = 28 etc.

Skip Counting by 5

In this method, we add the number 5 for each counting. In this manner, we get numbers with a difference of 5. For example, if we start counting from 5, then by skipping the count of numbers, we get the number series as

5,10,15,20,25,30,35,40,45,50,55,60,65,70…

Skip Counting by 10

When we skip count by 10, then we have to add 10 in each step. The procedure we follow for skip count 10 is

0+10=10, 10+10=20, 20+10=30, 30+10=40, 40+10=50, 50+10=60…

Skip Counting by 25

In this method, we add the number 25 for each counting. In this manner, we get numbers with a difference of 25. For example, if we start counting from 25, then by skipping the count of numbers, we get the number series as

25,50,75,100,125,150,175,200,225… etc

Skip Counting by 100

In this method, we add the number 100 for each counting. In this manner, we get numbers with a difference of 100. For example, if we start counting from 100, then by skipping the count of numbers, we get the number series as

100,200,300,400,500,600,700… etc

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In this article, we are going to introduce a new concept that is Cube of the Sum of Two Binomials terms. We are providing the different problems with a clear explanation on this topic. Follow our page and get full of knowledge on it. Firstly, to find the cube of the sum of two binomials, we need to multiply the binomials term three times. Refer to Solved Examples on Cube of Sum of Two Binomials provided along with Solutions for better understanding of the concept.

How to find the Cube of Sum of Two Binomials?

For example, (x + y) ^3 = (x + y) (x + y)^2 is the example of a binomial expression.

Here, we have an equation in an algebra like (a + b)^2 = a^2 + 2ab + b^2.
By using the above equation, we can expand the cube term.
(x + y) (x + y)^2 = (x + y) (x^2 + 2xy + y^2).
Multiply the terms (x + y) and (x^2 + 2xy + y^2). Then we get
(x + y) (x^2 + 2xy + y^2) = x (x^2 + 2xy + y^2) + y (x^2 + 2xy + y^2).
= x^3 + 2x^2y + xy^2 + yx^2 + 2xy^2 + y^3.
= x^3 + 3x^2y + 3xy^2 + y^3.
= x^3 +y^3 + 3xy(x + y).

Also, Read: Cube of a Binomial

Cube of Sum of Two Binomials Examples

1. Determine the expansion of (x + 2y)^3.

Solution:
The given expression is (x + 2y)^3.
We have an equation on cubes like (x + y)^3 = x^3 + y^3 + 3xy(x + y).
By comparing the above expression with the (x + y)^3
Here, x = x and y = 2y
Substitute the terms in the equation (x + y)^3
That is, (x + 2y)^3 = x^3 + (2y)^3 + 3x(2y)(x + 2y).
= x^3 + 8y^3 + 6xy(x + 2y).
= x^3 + 8y^3 + 6x^2y + 12xy^2.

Therefore, (x + 2y)^3 is equal to x^3 + 8y^3 + 6x^2y + 12xy^2.

2. Evaluate (55)^3.

Solution:
The given one is (55)^3.
We can write it as (50 + 5)^3.
(x + y)^3 = x^3 + y^3 + 3xy(x + y).
By comparing the (50 + 5)^3 with the above expression.
x = 50 and y = 5.
Substitute the values in the expression.
(50 + 5)^3 = (50)^3 + (5)^3 + 3(50)(5)(50 + 5).
= 1,25,000 + 125 + 750(55).
= 1,25,000 + 125 + 41,250.
= 1,66,375.

Therefore, (55)^3 is equal to 1,66,375.

3. Find the value of 64x^3 + y^3 if 4x + y = 6 and xy = 5.

Solution:
The given expression is 64x^3 + y^3.
4x + y = 6.
Cube the terms on both sides. Then, we will get
(4x + y)^3 = (6)^3.
We have an equation (x + y)^3 = x^3 + y^3 + 3xy(x + y).
Here, x = 4x and y = y.
Substitute the values in the equation. Then,
(4x)^3 + y^3 + 3(4x)(y)(4x + y) = 216.
64x^3 + y^3 + 12xy(4x + y) = 216.
But 4x + y = 6 and xy = 5.
So, 64x^3 + y^3 + 12(5)(6) = 216.
64x^3 + y^3 + 360 = 216.
64x^3 + y^3 = -144.

Therefore, 64x^3 + y^3 is equal to -144.

4. If a + 1/a = 3, find the values of a^3 – 1/a^3.

Solution:
The given term is a + 1/a = 3
Cube the terms on both sides. Then, we will get
(a + 1/a)^3 = (3)^3.
We have an equation (x + y)^3 = x^3 + y^3 + 3xy(x + y).
By comparing the given terms with the equation.
Here, x = a and y = 1/a.
By substituting the terms in the given equation. We will get
(a + 1/a)^3 = a^3 + (1/a)^3 + 3a(1/a)(a + 1/a) = 27.
= a^3 + 1/a^3 +3(a + 1/a) = 27.
We have (a + 1/a) = 3. By substituting this value in the above expression.
a^3 + 1/a^3 + 3(a + 1/a) =27.
a^3 + 1/a^3 + 3(3) = 27.
a^3 + 1/a^3 +9 = 27.
a^3 + 1/a^3 = 27 – 9 = 18.

Therefore, a^3 + 1/a^3 is equal to 18.

5. Expand the term (2x + y)^3.

Solution:
The given expression is (2x + y)^3.
we have an equation (x + y)^3 = x^3 + y^3 + 3xy(x + y).
by comparing the (2x + y)^3 with the above equation.
Here, x = 2x and y = y.
(2x + y)^3 = (2x)^3 + (y)^3 +3(2x)(y)(2x + y).
= 8x^3 + y^3 + 6xy(2x + y).
= 8x^3 + y^3 + 12x^2y + 6xy^2.

The final answer is 8x^3 + y^3 + 12x^2y + 6xy^2.

A Rational Number is represented as a fraction. For example, x / y is a rational number. Here, the upper term of the fraction is called a numerator that is ‘x’, and the lower term of the fraction is called a denominator that is ‘y’. Both terms numerator (x) and denominator (y) must be integers. Integer numbers are both positive and negative numbers like -4, -3, -2, -1, 1, 2, 3, 4…etc… The most important thing in the rational numbers is the denominator of the rational number cannot be equal to zero.

Rational Numbers are 2/3, 4/5, 6/7, 8/9, 10/15, 20/4, 125/100, ….etc… Check out the complete concept of Rational Numbers in Terminating and Non-Terminating Decimals below.

Rational Numbers in Decimal Fractions

By simplifying the rational numbers, we will get the result in the form of decimal fractions. We have two types of decimal fractions. They are

• Terminating Numbers
• Non – Terminating Numbers

When we converted the rational numbers into a decimal fraction, we will get either finite numbers of digits or infinite numbers of digits after the decimal point. If we get the decimal fraction with the finite number of digits, then it is called Terminating Numbers. If we get the decimal fraction with the infinite number of digits after the decimal point, then it is called Non – Terminating Numbers.

Terminating Number

Example for Terminating Numbers are 1.25, 0.68, 1.234, 2.456, 3. 4567, 5.687, 6.24, 8. 46, ….etc….The below examples are in the form of rational numbers and we need to convert that numbers into the form of decimal numbers.
(1) x / y = 100 / 25 = 0.4
(2) x / y = 644 / 8 = 8.5
(3) x / y = 5 / 4 = 1.25

Non – Terminating Number

Example for Non – Terminating Numbers are 1.23333, 2.566666, 5.8678888, 3.467777, 4.6899999,…..etc… The below-mentioned x / y fraction indicates the rational numbers and by simplifying it, we will get the decimal numbers.
(1) x / y = 256 / 6 =42.66666…
(2) x / y = 10 / 3 = 3.33333…
(3) x / y = 20 / 9 = 2.222222…

Note: If a rational number (≠ integer) can be expressed in the form p/(2^n × 5^m) where p ∈ Z, n ∈ W, and m ∈ W then the rational number will become a terminating decimal. If not, the rational number becomes the Non – Terminating Numbers.

Examples of Repeating and Non-Repeating Decimals

1. Find out the conversion of rational numbers to terminating decimal fractions?

(i) 1/4 is a rational fraction of form p/q. When this rational fraction is converted to decimal it becomes 0.25, which is a terminating decimal fraction.

(ii) 1/8 is a rational fraction of form p/q. When this rational fraction is converted to decimal fraction it becomes 0.125, which is also an example of a terminating decimal fraction.

(iii) 4/40 is a rational fraction of form p/q. When this rational fraction is converted to decimal fraction it becomes 0.1, which is an example of a terminating decimal fraction.

2. Find out the conversion of rational numbers to nonterminating decimal fractions.

(i) 1/11 is a rational fraction of form p/q. When we convert this rational fraction into a decimal, it becomes 0.090909… which is a non-terminating decimal.

(ii) 1/13 is a rational fraction of form p/q. When we convert this rational fraction into a decimal, it becomes 0.0769230769230… which is a non-terminating decimal.

(iii) 2/3 is a rational fraction of form p/q. When this is converted to a decimal number it becomes 0.66666667… which is a non-terminating decimal fraction.

Irrational Numbers

You may see different types of numbers such as real numbers, whole numbers, rational numbers, etc. Now, let us check out the irrational numbers. Irrational numbers are also real numbers that are represented as a simple fraction. There is no repeating or no terminate pattern available in Irrational Numbers. the numbers which do not consist of exact square roots of integers treats as Irrational Numbers. Also, the Irrational Number is pi and that is equal to the value of 3.14.

Solved Problems on Rational and Irrational Numbers

Addition

Add the two rational numbers. For example x / y = 1 / 2 and p / q = 2 / 6
To add the two rational numbers, we need to find out the LCM of the denominators.
That is, LCM of 2 and 6 is 6.
x / y + p / q = 1 / 2 + 2 / 6 = (3 + 2) / 6 = 5 / 6.

Multiplication

Multiply the two rational numbers such as 2 / 3 and 5 / 6.
x / y = 2 / 3 and p / q = 5 / 6
x / y X p / q = 2 / 3 x 5 / 6
(xX p) / (y X q) = (2 x 5) / 3 x 6)
px / yq = 10 / 18 = 5 / 9.

Subtraction

Subtract the two rational numbers. Here, 5 / 8 and 12 / 5 are rational numbers.
x / y = 5 / 8 and p / q = 12 / 5
x / y – p / q = 5 / 8 – 12 / 5
For subtraction, we need to find out the LCM of denominator values.
LCM of 8 and 5 is 40.
5 / 8 – 12 / 5 = [(5 x 5) – (12 x 8)] / 40 = (25 – 96) / 40 = -71 / 40.

Division

To divide the two rational numbers, we need to cross multiply the terms. For example, x / y and p / q are two rational numbers.
(x / y) ÷( p / q) = (2 / 5) ÷ ( 3 / 7)
Cross Multiply the first fraction numerator with second fraction denominator and vice versa.
xq / py = (2 x 7) / (5 x 3 )
xq / py= 14 / 15.

Everyone must know the number of Months and Days in a year. One complete year has 365 days or 366 days. Here, a leap year contains 366 days which comes once every four years. One year whether it is a leap year or a normal year contains 12 months. Now, months are divided into days that is 28, 29, 30, or 31 days.

Every year starts with the month of January and ends with the month of December. In between these two months, we have February, March, April, May, June, July, August, September, October, and November. In the below table, you can see the number of months in a year, the name of the month, the number of days in every month, and the short-form of every month.

List of Months and Days

Below is the list of Months and Days in a year. Learn the number of months and days along with their short-form whenever you need them. They are along the lines

S. No	Name of the month	Number of Days in a month	Shortform
1.	January	31	Jan
2.	February	28 or 29 (leap year)	Feb
3.	March	31	Mar
4.	April	30	Apr
5.	May	31	May
6.	June	30	Jun
7.	July	31	Jul
8.	August	31	Aug
9.	September	30	Sep
10.	October	31	Oct
11.	November	30	Nov
12.	December	31	Dec

Months and Days Examples

Now, calculate the number of days in a year based on the number of days in every month. That is, Count starts from the month of January
= (31 + 28 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + 31 + 30 + 31)
= 365 days for a common year.

Now, calculate the number of days for a leap year. That is,
= (31 + 29 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + 31 + 30 + 31)
= 366 days for a leap year.

Seven days are equal to 1 week. We need to divide the total number of days in a year by seven. That is,
365 / 7 = 52 weeks and 1 day.
And, for leap year
366 / 7 = 52 weeks and 2 days.
Every leap year is divided by four.

For example, 2000, 2004, 2008, 2012, 2016, 2020, 2024, and etc… are leap years. The remaining years, which are not divided by four are common years.

We have different seasons for every three months. That means December, January, and February comes under the winter season. March, April, and May considered as Spring Season. Next, the Summer season covers June, July, and August months. Finally, September, October, and November come under the autumn season. Totally four seasons in a year.

If two straight lines coincide at a point, then it is called the angle. The angle of elevation is defined as the angle between the horizontal plane and oblique line from the observer’s eye to an object which is located above his eye. Finally, the angle formed above the surface. It is a widely used concept and it is related to height and distance. The different terms used in the angle of elevation concept, trigonometric formulas, and solved examples are mentioned below for the sake of students.

Angle of Elevation – Definition

The angle of elevation of an object seen by the observer is the angle formed between the horizontal and the line from the object to the observer’s eye. The line in which the observer’s eye is there is known as the line of sight.

From the above image, you can see that the observer is looking at the object from the ground making an angle θ with the line of sight and horizontal line. By joining the imaginary line between the object and the end of the horizontal line, it forms a right-angle triangle. Thus we can use the trigonometry concept to find the distance from the observer to the object.

Example:

A student named Rakesh is looking at a bird sitting at the building, then there is an angle formed, which is inclined towards the bird from the Rakesh eye. This elevation angle is helpful to find the distance, height of the building with the trigonometric function like sine, cosine, and tangent.

Terms used in Elevation Angle

The three most important terms used in this concept are angle, horizontal line, and line of sight. Let us discuss each of them in detail in the following sections.

Horizontal Line:

A straight line on the flat surface where all points lie on a line and those points will have the same y-coordinate value. The angle and horizontal lines combine to form the angle of elevation.

Angle:

If two line segments meet at an endpoint, then the point is called the vertex. When two straight lines meet at a common point form an angle. It is also defined as the gap between two limes that connect on one side. It is measured in degrees or radians.

Line of Sight:

The line which is drawn from the observer’s eye to the object is known as the line of sight. In the angle of elevation, the object is kept above the line of sight of the observer. If you know the elevation angle, then it is easy to determine the distance and altitude.

Also, Check:

• Pairs of Angles
• Lines and Angles
• Angles

Angle of Elevation Formulas

Trigonometric ratios are used to find the unknown measures of the opposite, adjacent, and hypotenuse sides of the right angle.

The trigonometric ratios are

sin θ = \(\frac { Opposite side }{ Hypotenuse } =\frac { y }{ h } \)

cos θ = \(\frac { Hypotenuse }{ Adjacent side } =\frac { h }{ x } \)

tan θ = \(\frac { Opposite side }{ Adjacent side } =\frac { x }{ y } \)

Another type of problem is along the lines.

PQ = y, TQ = SR = I, PT = (y – l), QR = ST = x, PS = h

The formulas are

sin θ = \(\frac { PT }{ PS } =\frac { y – l }{ h } \)

cos θ = \(\frac { QR }{ PS } =\frac { x }{ h } \)

tan θ = \(\frac { PT }{ QR } =\frac { y – l }{ x } \)

Angle of Elevation Problems

Example 1:

A tower stands vertically on the ground. From a point on the ground, which is 48 m away from the foot of the tower, the angle of elevation of the top of the tower is 30°. Find the height of the tower.

Solution:

Let AB be the height of the tower.

Take AB = h and BC is the distance between the tower and the point C. In the right-angled trinagle ABC, ∠ACB = 30°

tan θ = \(\frac { AB }{ BC } \)

tan 30 = \(\frac { h }{ 48 } \)

\(\frac { 1 }{ √3 } \) = \(\frac { h }{ 48 } \)

cross multiply the fractions

48 = h√3

h = \(\frac { 48 }{ √3 } \)

h = 16√3

Therefore, the height of the tower is 16√3 m

Example 2:

A ladder of length 20 m is kept in against a wall of length 15 m such that their topmost point is in contact with one another and their bottom point is at a certain distance as shown in the figure. Find the angle subtended by the ladder on the floor.

Solution:

The length of the ladder is PQ = 20 m

The height of the wall is PR = 15 m

We have to find angle PQR = angle subtended by the ladder on the floor.

Let angle PQR = θ

We know that,

sin θ = \(\frac { PR }{ PQ } \)

sin θ = \(\frac { 15 }{ 20 } \)

θ = sin^-1 (\(\frac { 15 }{ 20 } \))

θ = sin^-1 (\(\frac { 3 }{ 4 } \))

θ = 48.59°

The angle subtended by the ladder on the floor is 48.59°.

Example 3:

A man of height 1 m 30 cm is standing in front of a tree of height 30 m. Find the angle of elevation to be made by the man’s eyes to look at the topmost point of the tree if the man is standing at a distance of 5 m from the tree.

Solution:

Here, PQ is the height of the tree = 30m

TR is the height of man = 1 m 30 cm = 1.30 m

RQ is the distance between the foot of the man and the tree = ST = 5 m

We have to find the angle of elevation, θ =?

We know that,

tan θ = \(\frac { (30 – 1.30) }{ 5 } \)

tan θ = 5.74

θ = tan^-1 (5.74)

θ = 80.117°

The angle of elevation to be made by the man’s eye is 80.117°.

Example 4:

When the angle of elevation of the Sum is 45°, the shadow of a coconut tree is 15 m in length. What is the height of the coconut tree?

Solution:

Let AB denote the height of the coconut tree and BC denotes the length of the shadow.

Therefore, according to the problem ∠ACB = 45°, BC = 18 m.

Let the height of the coconut tree AB = x meters.

Now, tan 45° = \(\frac { AB }{ BC } \)

1 = \(\frac { x }{ 18 } \)

x = 18

Therefore, the height of the coconut tree is 18 meters.

FAQs on Angle of Elevation

1. What is the angle of elevation in trigonometry?

The angle of elevation is the angle between the horizontal line of the sight and the line of sight up to an object.

2. Differences between the angle of elevation and depression?

The angle of depression is opposite to the angle of elevation. In the angle of depression, the observer is standing at the height and the object is placed below the line of sight of the observer.

3. What is the formula for Angle of Elevation?

The formula for Angle of Elevation is given by

sin θ = \(\frac { Opposite side }{ Hypotenuse } =\frac { y }{ h } \)

cos θ = \(\frac { Hypotenuse }{ Adjacent side } =\frac { h }{ x } \)

tan θ = \(\frac { Opposite side }{ Adjacent side } =\frac { x }{ y } \)

Know the probability of tossing three coins here. Follow the various terminology and methods involved in probability. There are various methods for tossing three coins. Check all the important formulae, fundamental counting theorem to find the probability in the later modules. Furthermore, you can get an idea of how to find the number of possible choices.  Also, find the solved examples of finding the probability of 3 coins with solutions.

Tossing of 3 Coins

When you toss 3 coins simultaneously, the possibility of outcomes are (HHT), or (TTH) or (HHH) or (THT) or (THH) or (HTH) or (HTT) or  (TTT), where H is called the heads and T is called the tails.

Therefore, the total number of toss for outcome = 2³=8

Important Formulae

• The possible outcomes of tossing 3 coins are {(HHT), or (TTH) or (HHH) or (THT) or (THH) or (HTH) or (HTT) or  (TTT)}
• The total no of possible outcomes = 8
• The probability of getting at least one head = number of possibilities of heads as outcome/total no of possibilities = 3/8
• The probability of getting at least two heads = number of possibilities of 2 heads as the outcome/total no of possibilities = 3/8
• The probability of getting no tail = number of possibilities of no tail as the outcome/total no of possibilities = 1/8

Fundamental Counting Theorem to find the Probability

To find the number of possible choices, the complete event of tossing coins is divided by the events of tossing each coin. The outcomes of the event of tossing the coin multiple times are similar to the event of tossing multiple coins.

Example: Tossing the coin for 4 times is similar to tossing 4 coins at once.

How to find the Number of Possible Choices?

When the coin has tossed, there are 2 possible outcomes, which shows either a tail or head.

The event is denoted with the letter (E)

For the Total Event

The total event is the event of tossing the first coins.

1st Sub-Event

The 1st event of tossing the first coins

This event can be defined in 2 ways

n(SE1) = 2

2nd Sub-Event

The 2nd event of tossing the second coin

This event can be defined in 2 ways

n(SE2) = 2

Total number of possible choices

n(E) = n(SE1) * n(SE2) * n(SE3)…

The sequence continues as many possibilities are present.

2 * 2 * 2 * 2…

2(number of coins)

Probabilities of Tossing Coins

• The probability of coin-flipping for 2 times and getting 3 tails in a row
• In case you flip the coin 2 times, finding the probability of getting exactly 3 tails.
• The probability of getting 3 tails while flipping 2 coins.
• The probability of getting exactly 3 tails when a coin is tossed 2 times.
• Finding the probability of getting at least 3 tails when a coin is tossed 2 times.

Probability of Tossing or Flipping Three Coins Examples

Problem 1:

When two of the unbiased coins are tossed, what is the probability of both showing head?

Solution:

As given in the question,

No of unbiased coins are tossed = 2

The probability of showing both heads is

P(E) = No of favourable outcomes/Total no of outcomes

Total no of outcomes is represented with sample space (S)

S = {HH, HT, TH, TT}

n(S) = 4

P(E) = 1/4 = 0.25

Problem 2:

Three coins are tossed, then find the probability of getting at least two heads?

Solution:

As given in the question,

No of coins = 3

Sample set (S) = {HHH, HTT, HHT, HTH, THH, THT, TTH, TTT}

The set of the favourable outcomes is denoted by E

E = the probability of getting at least two heads

E = { HHH, HHT, HTH, THH} = n(E) = 4

The probability of getting at least two heads = No of favourable outcomes/Total no of outcomes

P (at least two heads) = 4/8 = 1/2 = 0.5

Problem 3:

Three coins are tossed, then find the probability of getting at least one head?

Solution:

As given in the question,

No of coins = 3

Sample set (S) = {HHH, HTT, HHT, HTH, THH, THT, TTH, TTT}

The set of the favourable outcomes is denoted by E

E = the probability of getting at least one head

E = { HHH, HHT, HTH, THH, HTH, THH, THT, TTH} = n(E) = 7

The probability of getting at least two heads = No of favourable outcomes/Total no of outcomes

P (at least two heads) = 7/8

How to find the Probability of Tossing Coins?

To find the probability of tossing coins, we have to follow various steps. They are listed in the below fashion and you can follow them to arrive at the solution easily.

1. Analyse the question and check for the sample space.
2. Now, note down the sample space (Total no of outcomes)
3. Find the favourable part by following the question and note down the event space.
4. Use the definition of probability i.e., No of favourable outcomes/Total no of outcomes and substitute the values in it.
5. Hence, get the final result and know the probability of tossing the coins.

Various Events of Tossing 3 Coins

1. A coin is tossed 3 times or 3 coins are tossed at a time and the results or the final value is recorded. Find the probability of finding the events?

A fair coin is tossed thrice or 3 unbiased coins are tossed at a time.

The sample space for the event is

Sample set (S) = {HHH, HTT, HHT, HTH, THH, THT, TTH, TTT}

Therefore, n(S) = 8

a) Event of getting exactly one head

Let C be the event of getting exactly one head

Therefore, C = { HTT, THT, TTH}

n(C) = 3

From the definition, the probability of getting exactly one head

P(C) = n(C)/n(S) = 3/8

Thus, the probability of getting exactly one head = 3/8

b) Event of getting exactly two heads

Let A be the event of getting exactly two heads

Therefore, A = {HHT, HTH, THH}

n(A) = 3

From the definition, the probability of getting exactly two heads

P(A) = n(A)/n(S) = 3/8

Thus, the probability of getting exactly two heads = 3/8

c) Event of getting all heads

Let B be the event of getting all heads

Therefore, B = (HHH)

n(B) = 1

From the definition, the probability of getting all heads

P(B) = n(B)/n(S) = 1/8

Thus, the probability of getting all heads = 1/8

d) Event of getting two or more heads

Let X be the event of getting two or more heads

Therefore, X = { HHH, HHT, HTH, THH, HTH, THH, THT, TTH}

n(X) = 7

From the definition, the probability of two or more heads

P(X) = n(X)/n(S) = 7/8

Thus, the probability of getting two or more heads = 7/8

e) Event of getting at least 1 head

Let B be the event of getting at least 1 head

Therefore, B = (HTT, THT, TTH, HHT, HTH, THH, HHH)

n(B) = 7

From the definition, the probability of at least 1 head

P(B) = n(B)/n(S) = 7/8

Thus, the probability of getting at least = 7/8

f) Event of getting atmost 2 heads

Let Y be the event of getting atmost 2 heads

Therefore, B = (HTT, THT, TTH, HHT, HTH, THH)

n(Y) = 6

From the definition, the probability of getting atmost 2 heads

P(Y) = n(Y)/n(S) = 6/8

Thus, the probability of getting atmost 2 heads = 3/4

One of the most important concepts in a Mathematical journey is Even numbers and Odd numbers. Students who learn basic maths will start with even & odd numbers after arithmetic operations. These Even and odd numbers calculations are crucial for many students who are studying computer science engineering. Those students also practice the logic behind the even and odd number concept to code and generate an easy way of determining a given number is even or odd. Here, we have shared the List of Even and Odd Numbers Between 1 and 100 along with the definition and examples.

Even and Odd Numbers Between 1 and 100 Chart

Here is the chart for Even and Odd Numbers Between 1 and 100. Even numbers from 1 to 100 are marked in blue color and the rest of the numbers are odd numbers from 1 to 100. Check out the chart and find the given number is even or odd easily.

Now, we will see the definitions for both even and odd numbers with examples for a clear understanding of the concept:

What are Even Numbers?

The numbers that are divided by 2 evenly are called Even numbers. However, even numbers cannot have decimals. One of the common ways to check the number is even or not is when the last digit of the number is having 0, 2, 4, 6, or 8. In the mathematical form, we can find an even number by using ‘2n’, where n is an integer,

Example: Find the number 28 is an even number or not?

Solution: We know that the mathematic form is 2n for finding the even number, so

2n = 28
n = 28/2 = 14

Hence, 28 is divided by 2 so it is an even number.

What are Odd Numbers?

The definition of Odd numbers is any number that cannot be divided by two. In short, a number in form of 2k+1, where k ∈ Z (i.e. integers) are called odd numbers. These numbers are not evenly divided by 2, which implies there is some remainder left after division. For example, 1, 3, 5, 7, etc.

Example: Find 85 is an odd number or not?

Solution: We know 2k + 1 are the odd numbers

To find the number is odd or not, we have to check the unit place is having an odd number.

Hence, 85 is an Odd number.

List of Even Numbers Between 1 and 100

Below are the even numbers that exist between 1 and 100:

2             4             6             8           10

12           14           16           18           20

22           24           26           28           30

32           34           36           38           40

42           44           46           48           50

52           54           56           58           60

62           64           66           68           70

72           74           76           78           80

82           84           86           88           90

92           94           96           98           100

List of Odd Numbers Between 1 and 100

The following are the odd numbers from 1 to 100:

1             3             5             7             9

11           13           15           17           19

21           23           25           27           29

31           33           35           37           39

41           43           45           47           49

51           53           55           57           59

61           63           65           67           69

71           73           75           77           79

81           83           85           87           89

91           93           95           97           99

FAQs on Even and Odd Numbers between 1 and 100

1. Is 55 an odd number or even?

As we can see 55 number is even or odd in the list of an odd number above. If not, check the unit place of 55, if it consists of an odd number that is not evenly divisible by 2, then 55 is an odd number.

2. How do you find if a number is even or odd?

When the given number is divisible by 2 evenly then it’s an even number else odd number. If the number is bigger like thousands or millions, check the unit place. If the unit place contains an even number, then the bigger number is even else it’s an odd one.

3. Where can I get the list of even and odd numbers between 1 and 100?

You can get the even and odd numbers between 1 and 100 list elaborately from this page of a reliable website.

In the initial stages, you might feel Maths Problems complex and difficult? Once, you understand the formula you will no longer the concept or problems difficult. Finding the Fractional Part of a Whole Number may seem a bit difficult if you aren’t aware of the formula. However, the formula for finding a fraction of a whole number involves simple division and multiplication. Learn the step-by-step procedure on How to find a Fraction of a Whole Number and also check the solved examples provided for better understanding.

How to find a Fraction of a Whole Number?

Finding the Fraction of a Whole Number is the same as multiplying the number and fraction. This method is quite simple when it comes to whole numbers. However, to solve the problems you need to know about basic multiplication and division.

• Obtain the numbers.
• Multiply the whole number with the numerator of the fraction. The denominator remains unchanged during the entire process.
• Divide the product by the denominator of the fraction.

Solved Examples on Finding a Fraction of a Number

1. Find \(\frac { 2}{ 3 } \) of 22?

Solution:

To find \(\frac { 2}{ 3 } \) of 22 multiply the numerator 2 with the whole number 22 and then divide the product with denominator 3.

i.e. \(\frac { 2}{ 3 } \) . 22

= \(\frac { 2.22}{ 3 } \)

= \(\frac { 44}{ 3 } \)

2. Find \(\frac { 3}{ 5 } \) of 15?

Solution:

To find \(\frac { 3}{ 5 } \) of 15 multiply the numerator 3 with the whole number 15 and then divide the product with denominator 5 i.e.

= \(\frac { 3}{ 5} \) . 15

= \(\frac { 3.15}{ 5 } \)

= \(\frac { 45}{ 5 } \)

= 9

3. Find \(\frac { 3 }{ 6 } \) of 54?

Solution:

To find \(\frac { 3}{ 6 } \) of 54 multiply the numerator 3 with the whole number 54 and then divide the product with denominator 6 i.e.

= \(\frac { 3}{ 6} \) . 54

= \(\frac { 3.54}{ 6} \)

= 27

Are you looking for ways on how to change Fractions? Then, take the help of this article and learn how to convert an improper fraction to a mixed number or whole number, from whole number or mixed number to improper fraction easily. Refer to Procedures on Changing Fractions to understand the concept better. Practice the Example Problems for Converting Improper Fractions to Mixed Number and Vice Versa.

How to Convert an Improper Fraction to Mixed Number?

To Change an Improper Fraction to Mixed Number follow the below-listed steps. They are as follows

• Divide the numerator with the denominator
• Note down the whole number
• Later make a note of the remainder as a numerator above the denominator

Example:

Convert \(\frac { 9 }{ 4 } \) to a Mixed Fraction?

Solution:

Step 1: Divide the Numerator with Denominator i.e. 9÷4 = 2 with a remainder of 1

Step 2: Write the Quotient obtained as the Whole Number. Note down the Remainder as Numerator in the Fraction while keeping the denominator of the fraction unchanged.

Step 3: Thus, we get 2 \(\frac { 1 }{ 4 } \)

Therefore, improper Fraction \(\frac { 9 }{ 4 } \) converted to Mixed Fraction is 2 \(\frac { 1 }{ 4 } \)

How to Convert a Mixed Number to Improper Fraction?

To convert a Mixed Number to an Improper Fraction go through the below-mentioned steps. They are as such

• Multiply the whole number part with the fraction’s denominator
• Add the result to the numerator.
• Later, place the result obtained in the earlier step as a numerator while keeping the denominator unaltered.

Example:

Convert 4 \(\frac { 3 }{ 5 } \) to Improper Fraction?

Solution:

Step 1: Multiply the whole number part with the denominator of the fraction i.e. 4*5 = 20

Step 2: Add the result to the numerator of the fraction i.e. 20+3 = 23

Step 3: Place the numerator value obtained in the earlier step and keep the denominator unaltered i.e. \(\frac { 23 }{ 5 } \)

Therefore Mixed Number \(\frac { 3 }{ 5 } \)  converted to Improper Fraction is \(\frac { 23 }{ 5 } \)

How to Convert a Whole Number to Improper Fraction?

Go through the following steps to change from a whole number to Improper Fraction and they are outlined as below

• To change a whole number to Improper Fraction simply place the whole number part as the numerator over the denominator 1.
• The resultant is called a reduced improper fraction.
• If you want to change it to an unreduced improper fraction you need to multiply the whole number with a fraction equivalent of 1.

Example

Convert 13 to Improper Fraction?

Solution:

In order to express Whole numbers as Improper Fractions you just need to place the whole number as the numerator over the denominator 1.

13 = \(\frac { 13 }{ 1 } \)

13 converted to Improper Fraction is \(\frac { 13 }{ 1 } \)

FAQs on Changing Fractions

1. What is an Improper Fraction?

An Improper Fraction is a Fraction in which the numerator is greater than the denominator.

2. What is meant by Mixed Fraction?

A Fraction represented with its quotient and remainder is called a Mixed Fraction or Mixed Number. In other words, the mixed fraction is a combination of a whole number and a proper fraction.

3. How to Convert an Improper Fraction into Mixed Number?

Divide the Numerator with the Denominator. Take the Quotient obtained as a Whole Number and the Remainder as Numerator in the Proper Fraction while keeping the Denominator Unchanged.

Do you want to expand trinomials easily without any confusion and hassle? You should refer to this page. Here, we have explained how to expand the Square of a Trinomial and perfect square trinomial definition and formulas. Students who need more subject knowledge about square trinomials and solve any kind of trinomial expansions must go with this article completely. In this article, you will also get some worked-out examples on Square of a Trinomial and Perfect square trinomial. So, let’s continue your read and learn the concept of square trinomial.

Perfect Square Trinomial Definition & Formula

An expression obtained from the square of the binomial equation is a perfect square trinomial. When the trinomial is in the form ax² + bx + c then it is said to be a perfect square, if and only if it meets the condition b² = 4ac.

The Perfect Square Trinomial Formula is as follows,

(ax)²+2abx+b² = (ax+b)²
(ax)²−2abx+b² = (ax−b)²

How to Expand the Square of a Trinomial?

Here, we are discussing the expansion of the square of a trinomial (a + b + c).

Let (b + c) = x

(i) Then (a + b + c)² = (a + x)² = a² + 2ax + x²
= a² + 2a (b + c) + (b + c)²
= a² + 2ab + 2ac + (b² + c² + 2bc)
= a² + b² + c² + 2ab + 2bc + 2ca

Therefore, (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca.

(ii) (a – b – c)² = [a + (-b) + (-c)]²
= a² + (-b²) + (-c²) + 2 (a) (-b) + 2 (-b) (-c) + 2 (-c) (a)
= a² + b² + c² – 2ab + 2bc – 2ca

Therefore, (a – b – c)² = a² + b² + c² – 2ab + 2bc – 2ca.

(iii) (a + b – c)² = [a + b + (-c)]²
= a² + b² + (-c)² + 2ab + 2 (b) (-c) + 2 (-c) (a)
= a² + b² + c² + 2ab – 2bc – 2ca

Therefore, (a + b – c)² = a² + b² + c² + 2ab – 2bc – 2ca.

(iv) (a – b + c)² = [a + (- b) + c]²

= a² + (-b²) + c² + 2 (a) (-b) + 2 (-b) (-c) + 2 (c) (a)
= a² + b² + c² – 2ab – 2bc + 2ca

Therefore, (a – b + c)² = a² + b² + c² – 2ab – 2bc + 2ca.

Solved Examples on Square Trinomial

1. Expand (x+4y+6z)²

Solution:

Given trinomial expression is (x+4y+6z)²

We know that (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca.

Here a=1x, b=4y, c=6z

Now, substitute the values in the expression of (a + b + c)²

Then (x+4y+6z)² = (1x)² + (4y)² + (6z)² + 2(1x)(4y) + 2(4y)(6z) + 2(6z)(1x)

= x² + 16y² + 36z² + 8xy + 48yz + 12zx

Hence, (x+4y+6z)² = x² + 16y² + 36z² + 8xy + 48yz + 12zx. 

2. Is the trinomial x² – 6x + 9 a perfect square?

Solution:

Given trinomial is x² – 6x + 9, now calculate the expression and find it is a perfect square or not.

x² – 6x + 9 = x² – 3x – 3x + 9
= x(x – 3) – 3(x – 3)
= (x – 3)(x – 3)

Otherwise,

x² – 6x + 9 = x² – 2(3)(x) + 3² = (x – 3)²

The factors of the given equation are a perfect square.

Therefore, the given trinomial is a perfect square.

3. Simplify a + b + c = 16 and ab + bc + ca = 40. Find the value of a² + b² + c².

Solution:

As per the given question, a + b + c = 16

Now, by squaring both sides, we get

(a+ b + c)² = (16)²

a² + b² + c² + 2ab + 2bc + 2ca = 256

a² + b² + c² + 2(ab + bc + ca) = 256

a² + b² + c² + 2 × 40 = 256 [Given, ab + bc + ca = 40]

a² + b² + c² + 80 = 256

At this step, we have to subtract 80 from both sides

a² + b² + c² + 80 – 80 = 256 – 80

a² + b² + c² = 176

Hence, the square of a trinomial formula will help us to expand and get the result for a² + b² + c² is 176. 

Trigonometric Identities are useful when you are dealing with Trigonometric Functions in an Algebraic Expression. Usually, the Trig Identities involve certain functions of one or more angles. There are Several Identities involving the angle of a triangle and side length.  Check out all Fundamental Trigonometric Identities derived from Trigonometric Ratios using Worksheet on Trigonometric Identities. Practice the List of Trigonometric Identities, their derivation, and problems easily taking the help of the Trig Identities Worksheet with Answers.

List of Trigonometric Identities

There are several Trigonometric Identities that are used while solving Trigonometric Problems. Have a glance at the basic or fundamental trigonometric identities listed below and make your job simple. They are as follows

Pythagorean Identities

• sin² a + cos² a = 1
• 1+tan² a  = sec²a
• cosec² a = 1 + cot² a

Reciprocal Identities

• Sin θ = \(\frac { 1 }{ Csc θ } \) or Csc θ = \(\frac { 1 }{ Sin θ } \)
• Cos θ = \(\frac { 1 }{ Sec θ } \) or Sec θ = \(\frac { 1 }{ Cos θ } \)
• Tan θ = \(\frac { 1 }{ Cot θ } \) or Cot θ = \(\frac { 1 }{ Tan θ } \)

Opposite Angle Identities

• Sin (-θ) = – Sin θ
• Cos (-θ) = Cos θ
• Tan (-θ) = – Tan θ
• Cot (-θ) = – Cot θ
• Sec (-θ) = Sec θ
• Csc (-θ) = -Csc θ

Complementary Angles Identities

• Sin (90 – θ) = Cos θ
• Cos (90 – θ) = Sin θ
• Tan (90 – θ) = Cot θ
• Cot ( 90 – θ) = Tan θ
• Sec (90 – θ) = Csc θ
• Csc (90 – θ) = Sec θ

Ratio Identities

• Tan θ = \(\frac { Sin θ }{ Cos θ } \)
• Cot θ = \(\frac { Cos θ }{ Sin θ } \)

Angle Sum and Difference Identities

Consider two angles , α and β, the trigonometric sum and difference identities are as follows:

• sin(α+β)=sin(α).cos(β)+cos(α).sin(β)
• sin(α–β)=sinα.cosβ–cosα.sinβ
• cos(α+β)=cosα.cosβ–sinα.sinβ
• cos(α–β)=cosα.cosβ+sinα.sinβ
• tan(α+β) = \(\frac { tanα+tanβ }{ 1-tanα.tanβ } \)
• tan(α-β) = \(\frac { tanα-tanβ }{ 1+tanα.tanβ } \)

Prove the following Trigonometric Identities

1. (1 – cos²θ) csc²θ  =  1?

Solution:

Let us consider L.H.S =  (1 – cos²θ) csc²θ  and  R.H.S  =  1.

L.H.S =  (1 – cos²θ) csc²θ

We know sin²θ + cos²θ  =  1,

sin²θ  =  1 – cos²θ

L.H.S =  sin²θ ⋅ csc²θ

We also know csc²θ =  1/ sin²θ

L.H.S =  sin²θ ⋅  1 / sin²θ

L.H.S = 1

Hence Proved, L.H.S = R.H.S

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2. Prove tan θ sin θ + cos θ  =  sec θ

Solution:

Let L.H.S  =  tan θ sin θ + cos θ  and R.H.S =  sec θ.

L.H.S =  tan θ sin θ + cos θ

We know tanθ = \(\frac { Sin θ }{ Cos θ } \)

L.H.S = \(\frac { Sin θ }{ Cos θ } \) ⋅ sin θ + cos θ

L.H.S =  Sin² θ / Cos θ+ cos θ

L.H.S = (sin²θ/cos θ) + (cos²θ/cosθ)

L.H.S = (sin²θ + cos²θ) / cos θ

L.H.S= \(\frac { 1}{ cos θ } \)

L.H.S = 1 / cos θ

= Sec θ

Therefore, L.H.S = R.H.S

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3. Prove cot θ + tan θ  =  sec θ csc θ?

Solution:

Let L.H.S  =  cot θ + tan θ and R.H.S  =  sec θ csc θ.

L.H.S =  cot θ + tan θ

L.H.S =  \(\frac { Cos θ }{ Sin θ } \) + \(\frac { Sin θ }{ Cos θ } \)

L.H.S = (cos²θ/sin θ cos θ) + (sin²θ/sin θ cos θ)

L.H.S = (cos²θ + sin²θ) / sin θ cos θ

L.H.S = \(\frac { 1 }{ sin θ cos θ } \)

L.H.S = \(\frac { 1 }{ cos θ } \)⋅ \(\frac { 1 }{ sin θ } \)

L.H.S =  sec θ csc θ

L.H.S = R.H.S

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4. Prove sec θ √(1 – sin²θ)  =  1?

Solution:

Let L.H.S  =  sec θ √(1 – sin²θ)  and R.H.S  =  1.

L.H.S =  sec θ √(1 – sin²θ)

We know sin²θ + cos²θ  =  1, we have

cos²θ  =  1 – sin²θ

Then,

L.H.S  =  sec θ √cos²θ

L.H.S =  sec θ ⋅ cos θ

L.H.S =  sec θ ⋅ \(\frac { 1 }{ sec θ } \)

L.H.S = \(\frac { sec θ }{ sec θ } \)

L.H.S =  1

L.H.S =  R.H.S

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5. Prove (1 – cos θ)(1 + cos θ)(1 + cot²θ)  =  1

Solution:

Let L.H.S  =  (1 – cos θ)(1 + cos θ)(1 + cot²θ)  =  1 and R.H.S  =  1.

L.H.S =  (1 – cos θ)(1 + cos θ)(1 + cot²θ)

L.H.S =  (1 – cos²θ)(1 + cot²θ)

We know sin²θ + cos²θ  =  1, we have

sin²θ  =  1 – cos²θ

Then,

L.H.S =  sin²θ ⋅ (1 + cot²θ)

L.H.S =  sin²θ  + sin²θ ⋅ cot²θ

L.H.S =  sin²θ  + sin²θ ⋅ (cos²θ/sin²θ)

L.H.S =  sin²θ + cos²θ

L.H.S =  1

Therefore, L.H.S = R.H.S

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6. Prove tan⁴θ + tan²θ  =  sec⁴θ – sec²θ?

Solution:

Let L.H.S  =  tan⁴θ + tan²θ  and R.H.S  =  sec⁴θ + sec²θ.

L.H.S =  tan⁴θ + tan²θ

L.H.S =  tan²θ (tan²θ + 1)

We know that,

tan²θ  =  sec²θ – 1

tan²θ + 1  =  sec²θ

Then,

L.H.S =  (sec²θ – 1)(sec²θ)

L.H.S  =  sec⁴θ – sec²θ

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7. Prove √{(sec θ – 1)/(sec θ + 1)}  =  cosec θ – cot θ?

Solution:

Let L.H.S  = √{(sec θ – 1)/(sec θ + 1)} and R.H.S  =  cosec θ – cot θ

= √{(sec θ – 1)/(sec θ + 1)}

= √[{(sec θ – 1) (sec θ – 1)}/{(sec θ + 1) (sec θ – 1)}]

= √{(sec θ – 1)² / (sec²θ – 1)}

= √{(sec θ – 1)² / tan²θ}

= (sec θ – 1)/tan θ

=  (sec θ/tan θ) – (1/tan θ)

=  {(1/cos θ)/(sin θ/cos θ)} – cot θ

= {(1/cos θ) ⋅ (cos θ/sin θ)} – cot θ

=  (1/sin θ) – cot θ

= cosec θ – cot θ

Therefore, L.H.S = R.H.S

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Wondering what are the Terms used in Division mathematical operation? Then, this page will explain to you all about Division Terms along with some solved examples. Students and Teachers who are searching for a clear explanation about the Terms of Division can make their look at this guide and understand the Division operation so easily. Let’s start with the definition of Division and then we will discuss the Terms used in Division with a brief explanation and worked-out problems.

Also Check: Properties of Division of Integers

Definition of Division

In maths, Division is a process of distributing a group of things into equal parts. Also, the division is one of the basic arithmetic operations that provides a fair output of sharing. In other words, the division is also a repeated subtraction.

However, the division is an operation inverse of multiplication. Mathematical Notation for Division is ÷, /. For example 13 ÷ 5 or 5/7.

Each and every part concerned in the Division Equation has special names.

Dividend ÷ Divisor = Quotient & Remainder

Let’s see the explanation of each part involved in the division equation from the below module.

Terms Used in Division – Explanation

There are four terms that represent the four numbers in a division sum. They are as such,

Dividend: Number that is being divided in the division method.
Divisor: Number that divides the dividend exactly.
Quotient: Number that results after the completion of the division process.
Remainder: “Left Over Number” after the division of the given number.

Facts About Division

• Division by zero (0) is undefined.
• The division of the same numerator and denominator is always One (1). For instance: 5/5 = 1.
• If you divide any number by one then the answer will be the original number. It means, when the divisor is 1 then the quotient will be equal to the dividend. For example: 20 ÷ 1= 20.

Solved Examples of Division Terms

1. Divide 65 by 5 & identify the Terms used in Division

Solution:

Here, the division process is shown in an image along with the division terms. So check out the image and get to know the terms of division:

Here, the final values of division terms are:

Dividend = Divisor × Quotient + Remainder ⇒ 65 = 5 × 13 + 0.

2. 20 Divided by 5

Solution:

First, divide 20 by 5 and then know which value is referred to which division term,

20 ÷ 5 = 4 ,

Here, 20 is Dividend

5 is Divisor

4 is Quotient

0 is Remainder ( Because there is no leftover number after the division process).

3. Divide 49474 by 7

Solution:

Given division expression is \(\frac { 49474 }{ 7 } \)

Now, we have to divide the 49474 by 7 and get the result.

After the division process, we get the outcome as \(\frac { 49474 }{ 7 } \) = 7067 R 5.

Here, the terms used in the division of \(\frac { 49474 }{ 7 } \) are 49474 is the dividend, 7 is the divisor, 7067 is the quotient and 5 is the remainder of the division.

For more division problems click on the Examples on Division of Integers and solve all the problems easily.