## RS Aggarwal Class 9 Solutions Chapter 4 Lines and Triangles Ex 4A

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 4 Lines and Triangles Ex 4A.

**Other Exercises**

RS Aggarwal Solutions Class 9 Chapter 4 Lines and Triangles Ex 4A

RS Aggarwal Solutions Class 9 Chapter 4 Lines and Triangles Ex 4B

RS Aggarwal Solutions Class 9 Chapter 4 Lines and Triangles Ex 4C

RS Aggarwal Solutions Class 9 Chapter 4 Lines and Triangles Ex 4D

**Question 1.**

**Solution:**

**(i)Angle :** When two rays OA and OB meet at a point o, then ∠AOB is called an angle.

**(ii) Interior of angle :** The interior an angle is a set of all points in its plane which lie on the same side of OA as B and also on the same side of OB as A.

**(iii) Obtuse angle :** An angle greater than 90° but less than 180° is called an obtuse angle.

**(iv) Reflex angle :** An angle more than 180° but less than 360° is called a reflex angle.

**(v) Complementary angles :** Two angles are said to be complementary angles if their sum is 90°.

**(vi) Supplementary angles :** Two angles are said to be supplementary angles if their sum is 180°.

**Question 2.**

**Solution:**

∠A = 36°27’46”

∠B = 28° 43’39”

Adding, ∠A + ∠B = 64° 70′ 85″

We know that 60″ = 1′ and 60′ = 1°

∠A+ ∠B = 65° 11′ 25″ Ans.

**Question 3.**

**Solution:**

36° – 24° 28′ 30″

= 35° 59’60” – 25° 28’30”

= 10° 31′ 30″ Ans.

**Question 4.**

**Solution:**

We know that two angles are complementary of their sum is 90°. Each of these two angles is complement to the other, therefore.

(i) Complement of 58° = 90° – 58° = 32°

(ii) Complement of 16° = 90° – 16° = 74°

(iii) Complement of \(\frac { 2 }{ 3 } \) of a right angle i.e.

of \(\frac { 2 }{ 3 } \) x 90° or 60° = 90° – 60° = 30°

= \(\frac { 2 }{ 3 } \) of right angle,

(iv) Complement of 46° 30′

= 90° – 46° 30′

= 43° 30′

(v) Complement of 52° 43′ 20°= 90° – 52° 43′ 20″

= 37° 16′ 40″

(vi) Complement of 68° 35′ 45″

= 90° – 68° 35′ 45″

= 21° 24′ 15″ Ans.

**Question 5.**

**Solution:**

We know that two angles are said to be supplement to each other of their sum is 180° therefore

(i) Supplement of 68° = 180° – 68° =112°

(ii) Supplement of 138° = 180° – 138° = 42°

(iii) Supplement of \(\frac { 3 }{ 5 } \) of a right angle or \(\frac { 3 }{ 5 } \) x 90° or 54°

= 180° – 54° = 126°

(iv) Supplement of 75° 36′ = 180° – 75° 36′ = 104° 24′

(v) Supplement of 124° 20′ 40″

= 180° – 124° 20′ 40″

= 55° 39′ 20″

(vi) Supplement of 108° 48′ 32″ = 180° – 108″ 48′ 32″ = 71° 11′ 28″ Ans.

**Question 6.**

**Solution:**

(i) Let the measure of required angle = x ,

their its complement = 90° – x

According to the condition,

x = 90° – x => 2x = 90°

=>x = \(\frac { { 90 }^{ o } }{ 2 } \) = 45°

Required angle = 45°

(ii) Let the measure of required angle = x then its supplement = 180° – x

According to the condition,

x = 180° – x => 2x = 180° = 90°

=>x = \(\frac {{ 180 }^{ o }}{ 2 }\) = 90°

Hence required angle = 90° Ans.

**Question 7.**

**Solution:**

Let required angle = x

then its complement = 90° – x

According to the condition,

x – (90° – x) = 36°

=> x – 90° + x = 36°

=> 2x = 36° + 90° = 126°

= \(\frac { { 126 }^{ o } }{ 2 } \) = 63°

Required angle = 63° Ans.

**Question 8.**

**Solution:**

Let the required angle = x

then its supplement = 180° – x

According to the condition,

(180° – x) – x = 25°

=> 180° – x – x = 25°

=> – 2x = 25° – 180°

=> – 2x = – 155°

=> x = \(\frac { { – 155 }^{ o } }{ – 2 } \)

= 77.5°

Hence required angle = 77.5° Ans.

**Question 9.**

**Solution:**

Let required angle = x

Then its complement = 90° – x

According to the condition,

x = 4 (90° – x) => x = 360° – 4x

=> x + 4x = 360° => 5x = 360°

x = \(\frac { { 360 }^{ o } }{ 5 } \) = 72°

Required angle = 72° Ans.

**Question 10.**

**Solution:**

Let required angle = x

Then its supplement = 180° – x

According to the condition,

x = 5 (180° – x)

=> x = 900° – 5x

=> x + 5x = 900°

=> 6x = 900°

=> x = \(\frac { { 900 }^{ o } }{ 6 } \) = 150°

Hence, required angle = 150° Ans

**Question 11.**

**Solution:**

Let required angle = x

then its supplement = 180° – x

and complement = 90° – x

According to the condition,

180° – x = 4 (90°- x)

=> 180° – x = 360° – 4x

=> – x + 4x — 360° – 180°

=> 3x= 180°

=> x = \(\frac { { 180 }^{ o } }{ 3 } \) = 60°

Required angle = 60° Ans.

**Question 12.**

**Solution:**

Let required angle = x

Then, its complement = 90° – x

and its supplement = 180° – x

According to the condition,

90° – x = \(\frac { 1 }{ 3 } \) (180° – x)

=> 90° – x = 60° – \(\frac { 1 }{ 3 } \) x

=> 90° – 60° = x – \(\frac { 1 }{ 3 } \) x

=> \(\frac { 2 }{ 3 } \) x = 30° =>x = \(\frac { { 30 }^{ o }X3\quad \quad }{ 2 } \) => x = 45° Ans.

**Question 13.**

**Solution:**

Let one angle = x

Then, its supplement = 180° – x

According to the condition,

x : (180° – x) = 3:2

=> \(\frac { x }{ { 180 }^{ o }-x } =\frac { 3 }{ 2 } \)

=>2x = 3(180°- x)

=> 2x = 540° – 3x

=> 2x + 3x = 540°

=> 5x = 540° => x = \(\frac { { 540 }^{ o } }{ 5 } \) = 108°

Angle = 108° and its supplement = 180° – 108° = 72°

Hence, angles are 108° and 72° Ans.

**Question 14.**

**Solution:**

Let angle = x

Then, its complementary angle = 90° – x

According to the condition,

x : (90° – x) = 4 : 5

=> \(\frac { x }{ { 90 }^{ o }-x } =\frac { 4 }{ 5 } \)

=> 5x = 4 (90° – x)

=> 5x = 360° – 4x

=> 5x + 4x = 360°

=> 9x = 360°

=> x = \(\frac { { 360 }^{ o } }{ 9 } ={ 40 }^{ o } \)

and its complement = 90° – 40° = 50°

Hence, angles are 40° and 50° Ans.

**Question 15.**

**Solution:**

Let the required angle = x

.’. its complement = 90° – x

and supplement = 180° – x

According to the condition,

7(90° – x) = 3(180° – x) – 10°

=> 630° – 7x = 3 (180° – x) – 10°

=> 630° – 7x = 540° – 3x – 10°

=> – 7x + 3x = 540° – 10° – 630°

– 4x = – 100°

x = \(\frac { { -100 }^{ o } }{ -4 } ={ 25 }^{ o }\)

Hence required angle = 25° Ans.

Hope given RS Aggarwal Solutions Class 9 Chapter 4 Lines and Triangles Ex 4A are helpful to complete your math homework.

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