Class 9th Math Exercise 6.1 | Example 1,2 and 3| Lines & Angles | NCERT

 

Example 1

In Fig. 6.9, lines PQ and RS intersect each other at point O. If ∠POR : ∠ROQ = 5 : 7, find all the angles.

Given:

∠POR : ∠ROQ = 5 : 7

Let,

∠POR = 5x

∠ROQ = 7x

Since ∠POR and ∠ROQ form a linear pair,

5x + 7x = 180°

12x = 180°

x = 15°

Therefore,

∠POR = 5 × 15° = 75°

∠ROQ = 7 × 15° = 105°

By vertically opposite angles,

∠QOS = ∠POR = 75°

∠POS = ∠ROQ = 105°

Hence, all the angles are:

∠POR = 75°

∠ROQ = 105°

∠QOS = 75°

∠POS = 105°

Example 2:

In Fig. 6.10, ray OS stands on a line POQ. Ray OR and ray OT are angle bisectors of ∠POS and ∠SOQ, respectively. If ∠POS = x, find ∠ROT.

Solution:

Given:

∠POS = x

Since OS stands on the straight line POQ,

∠POS + ∠SOQ = 180°

x + ∠SOQ = 180°

∠SOQ = (180° − x)

Now, OR bisects ∠POS,

∠ROS = x/2

Also, OT bisects ∠SOQ,

∠SOT = (180° − x)/2

Therefore,

∠ROT = ∠ROS + ∠SOT

= x/2 + (180° − x)/2

= 180°/2

= 90°

Hence,

∠ROT = 90°

Example 3:

In Fig. 6.11, OP, OQ, OR and OS are four rays. Prove that

∠POQ + ∠QOR + ∠SOR + ∠POS = 360°.

Solution :

 Given

OP, OQ, OR and OS are four rays meeting at point O.

We know that the sum of all angles around a point is 360°.

The four angles around point O are:

∠POQ, ∠QOR, ∠SOR and ∠POS

Therefore,

∠POQ + ∠QOR + ∠SOR + ∠POS = 360°

Hence proved.