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# Class 9 RD SHARMA Solutions Maths Chapter 10 - Lines and Angles

## Lines and Angles Exercise Ex. 10.1

### Solution 2

(i) 54°

Since, the sum of an angle and its supplement is 180°

Its supplement will be 180° - 54° = 126°.

(ii) 132°

Since, the sum of an angle and its supplement is 180°

Its supplement will be 180° - 132° = 48°.

(iii) 138°

Since, the sum of an angle and its supplement is 180°

Its supplement will be 180° - 138° = 42°.

### Solution 13

Let the measure of the angle be xo.

Its complement will be (90o - xo) and its supplement will be (180o - xo).

Supplement of thrice of the angle = (180o - 3xo)

According to the given information:

(90o - xo) = (180o - 3xo)

3x - x = 180 - 90

2x = 90

x = 45

Thus, the measure of the angle is 45o.

The measure of the angle is 45o

## Lines and Angles Exercise Ex. 10.2

### Solution 19

Given that OR ⊥ PQ

POR = 90o

⇒ POS  + SOR = 90o

ROS = 90 - POS                ... (1)

QOR = 90o                     (As OR ⊥ PQ)

QOS - ROS = 90o

ROS = QOS - 90o             ... (2)

On adding equations (1) and (2), we have

2 ROS = QOS - POS

(i) True

(ii) False

(iii) False

(iv) True

(i) obtuse.

(ii) 180o

(iii) uncommon

## Lines and Angles Exercise Ex. 10.4

### Solution 8

Let AB and CD be perpendicuar to line MN.

### Solution 11

Let AB and CD be perpendicuar to line MN.

### Solution 23

Consider the angles AOB and ACB.

## Lines and Angles Exercise 10.56

### Solution 25

Correct option (c)

Let one angle be θ

Then, its complementary = 90 - θ

According to question,

2θ = 3(90 - θ)

5θ = 270

θ = 54°

Then,  90 - θ° = 36°

Hence, the smaller angle is 36°.

Hence, correct option is (c).