RS Aggarwal Solutions Class 9 Chapter 7 Exercise 7A

Get step by step solutions of RS Aggarwal Class 9 Chapter 7 Exercise 7A (Lines and Angles)

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RS Aggarwal Veena Aggarwal Class 9 Chapter 7 Ex. 7A Solution

1: Define the following terms:

(i) Angle
When two rays originated from same point, then an angle is formed.

(ii) Interior of an angle
The area between two rays which is originated from same point, is called Interior of an angle.

(iii) Obtuse angle
An angle which is measured more than 90⁰ but less than 180⁰ is called Obtuse angle.

(iv) Reflex angle
An angle which is measured more than 180⁰ but less than 360⁰ is called Reflex angle.

(v) Complementary angle
Two angle are said to be complementary if, the sum of both angles are 90⁰ .

(vi) Supplementary angle
Two angle are said to be supplementary, if the sum of both angles are 180⁰.

2: Find the complement of each of the following angles:

(i) 55⁰ (ii) 16⁰ (iii) 90⁰ (iv) 2/3 of a right angle

Complement of 55⁰ = 90⁰ – 55⁰ = 35⁰

Complement of 16⁰ = 90⁰ – 16⁰= 74⁰

Complement of 90⁰ = 90⁰ – 90⁰ = 0⁰

Complement of 2/3 of a right angle
= 90⁰ – ( 90⁰ × 2/3) = 90⁰ – 60⁰ =30⁰

3: Find the supplementary of each of the following angles:

(i) 42⁰      (ii) 90⁰       (iii) 124⁰    (iv) 3/5 of a right angle

Supplementary of 42⁰ = 180⁰ – 42⁰ =138⁰

Supplementary of 90⁰ = 180⁰ – 90⁰ = 90⁰

Supplementary of 124⁰ = 180⁰ –  124⁰ = 56⁰

Supplementary of 3/5 of a right angle
= 180⁰ – ( 90⁰ × 3/5) = 180⁰ – 54⁰= 126⁰

4: Find the measure of an angle which is

(i) equal to its complement,
Let the angle is x⁰ , then
⇒ x⁰ = 90⁰ – x⁰
⇒ 2x⁰ = 90⁰
⇒ x⁰ = 45⁰

Hence the angle is 45⁰ which is equal to its complement.

(ii) equal to its supplement
Let the angle is x⁰ then
⇒ x⁰ = 180⁰ – x⁰
⇒ 2x⁰ = 180⁰
⇒ x⁰ = 90⁰

Hence the angle is 90⁰ which is equal to its supplement.

5: Find the measure of an angle which is 36⁰ more than its complement.

Let the angle is x⁰ , then its complement angle will be 90⁰ – x⁰
According to question,
⇒ x⁰ – 36⁰ = 90⁰ – x⁰
⇒ 2x⁰ = 90⁰ + 36⁰ = 126⁰
⇒ x⁰ = 63⁰

Hence the angle is 63⁰ .

6: Find the measure of an angle which is 30⁰ less than its supplement.

Let the angle is x⁰, then its supplement angle will be 180⁰ – x⁰

According to question,
⇒ x⁰ +30⁰ = 180⁰ – x⁰
⇒ 2x⁰ = 180⁰ – 30⁰ = 150⁰
⇒ x⁰ = 75⁰
Hence the angle is 75⁰.

7: Find the angle which is four times its complement.

Let the angle is x⁰, then its complement angle will be 90⁰ – x⁰
According to question,
⇒ x⁰ = 4(90⁰ – x⁰ )
⇒ x⁰ = 360⁰ – 4x⁰
⇒ 5x⁰ = 360⁰
⇒ x⁰= 72⁰

Hence the angle is 72⁰

8: Find the angle which is five times its supplement.

Let the angle is x⁰, then its supplementary angle will be 180⁰ – x⁰
According to question,
⇒ x⁰ = 5(180⁰ – x⁰ )
⇒ x⁰ = 900⁰ – 5x⁰
⇒ 6x⁰ = 900⁰
⇒ x⁰ = 150⁰

Hence the angle is 150⁰ .

9: Find the angle whose supplement is four times its complement.

Let the angle is x0, then its supplement angle will be 180⁰ – x⁰
and complement will be 90⁰ – x⁰

According to question,
⇒ (180⁰ – x⁰ ) = 4(90⁰ – x⁰ )
⇒ 180⁰ – x⁰ = 360⁰ – 4x⁰
⇒ 4x⁰ – x⁰ = 360⁰ – 180⁰ = 180⁰
⇒ 3x⁰=180⁰
⇒ x⁰ = 60⁰

10: Find the angle whose complement is one third of its supplement.

Let the angle is x⁰ , then its supplement angle will be 180⁰ – x⁰
and complement will be 90⁰ – x⁰

According to question,
⇒ (90⁰ – x⁰) = 1/3 (180⁰ – x⁰)
⇒270⁰ – 3x⁰= 180⁰ – x⁰
⇒ – 3x⁰ + x⁰ = 180⁰ – 270⁰ = –90⁰
⇒ – 2x⁰ = – 90⁰
⇒ x⁰ = 45⁰

Hence the angle is 45⁰ .

11: Two complementary angles are in the ratio 4 : 5. Find the angles.

Hence the angles are 40⁰ and 50⁰ .

12: Find the value of x for which the angles (2x – 5)⁰
and (x – 10)⁰ are complementary angles.

Since the given angle are complementary to each other, therefore sum of both angle will be 90⁰ ,

So, (2x – 5)⁰ + (x – 10)⁰ = 90⁰
⇒ (3x – 15)⁰ = 90⁰
⇒3x = 90⁰ +15⁰ = 105⁰
⇒ x⁰ = 35⁰
Hence the angle is 35⁰.