RD Sharma Class 9 Solutions Chapter 8 Exercise 8.1 (Updated for 2023)

RD Sharma Class 9 Solutions Chapter 8 Exercise 8.1: RD Sharma Solutions for Class 9 Maths are given here for Chapter 8 – Lines and Angles. In order to make concepts easier for students, we have provided RD Sharma Class 9 Maths Chapter 8 solutions to help them go through several solved exercises and, at the same time, practice the questions. RD Sharma Solutions for Class 9 can help students discover new ways to solve difficult problems.

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RD Sharma Class 9 Solutions Chapter 8 Exercise 8.1

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Question 1: Write the complement of each of the following angles:

(i)20⁰

(ii)35⁰

(iii)90⁰

(iv) 77⁰

(v)30⁰

Solution:

(i) The sum of an angle and its complement = 90⁰

Therefore, the complement of 20⁰ = 90⁰ – 20⁰ = 70⁰

(ii) The sum of an angle and its complement = 90⁰

Therefore, the complement of 35° = 90° – 35° = 55

(iii) The sum of an angle and its complement = 90⁰

Therefore, the complement of 90⁰ = 90⁰ – 90⁰ = 0⁰

(iv) The sum of an angle and its complement = 90⁰

Therefore, the complement of 77⁰ = 90° – 77⁰ = 13⁰

(v) The sum of an angle and its complement = 90⁰

Therefore, the complement of 30⁰ = 90⁰ – 30⁰ = 60⁰

Question 2: Write the supplement of each of the following angles:

(i) 54⁰

(ii) 132⁰

(iii) 138⁰

Solution:

(i) The sum of an angle and its supplement = 180⁰.

Therefore supplement of angle 54⁰ = 180⁰ – 54⁰ = 126⁰

(ii) The sum of an angle and its supplement = 180⁰.

Therefore supplement of angle 132⁰ = 180⁰ – 132⁰ = 48⁰

(iii) The sum of an angle and its supplement = 180⁰.

Therefore supplement of angle 138⁰ = 180⁰ – 138⁰ = 42⁰

Question 3: If an angle is 28⁰ less than its complement, find its measure.

Solution:

Let the measure of any angle is ‘ a ‘ degrees

Thus, its complement will be (90 – a) ⁰

So, the required angle = Complement of a – 28

a = ( 90 – a ) – 28

2a = 62

a = 31

Hence, the angle measured is 31⁰.

Question 4: If an angle is 30° more than one-half of its complement, find the measure of the angle.

Solution:

Let an angle measured by ‘ a ‘ in degrees

Thus, its complement will be (90 – a) ⁰

Required Angle = 30⁰ + complement/2

a = 30⁰ + ( 90 – a ) ⁰ / 2

a + a/2 = 30⁰ + 45⁰

3a/2 = 75⁰

a = 50⁰

Therefore, the measure of the required angle is 50⁰.

Question 5: Two supplementary angles are in the ratio 4:5. Find the angles.

Solution:

Two supplementary angles are in the ratio 4:5.

Let us say, the angles are 4a and 5a (in degrees)

Since angles are supplementary angles;

Which implies, 4a + 5a = 180⁰

9a = 180⁰

a = 20⁰

Therefore, 4a = 4 (20) = 80⁰ and

5(a) = 5 (20) = 100⁰

Hence, the required angles are 80° and 100⁰.

Question 6: Two supplementary angles differ by 48⁰. Find the angles?

Solution: Given: Two supplementary angles differ by 48⁰.

Consider a⁰ to be one angle then its supplementary angle will be equal to (180 – a) ⁰

According to the question;

(180 – a ) – x = 48

(180 – 48 ) = 2a

132 = 2a

132/2 = a

Or a = 66⁰

Therefore, 180 – a = 114⁰

Hence, the two angles are 66⁰ and 114⁰.

Question 7: An angle is equal to 8 times its complement. Determine its measure.

Solution: Given: Required angle = 8 times of its complement

Consider a⁰ to be one angle then its complementary angle will be equal to (90 – a) ⁰

According to the question;

a = 8 times its complement

a = 8 ( 90 – a )

a = 720 – 8a

a + 8a = 720

9a = 720

a = 80

Therefore, the required angle is 80⁰.

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