Question

In Fig. 6.17, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that ROS = ½ (QOS – POS).​

Answer 1

Answer:

You didnt send the figure, but i found the figure (see file attached below)

Now,

using linear pair, ∠QOR+∠ROS+∠POS=180°

∠ROS+∠POS+90°=180°...........∠QOR=90°

∠ROS+∠POS=90°.......(1)

Now,

∠QOS=∠ROS+∠ROQ

∠QOS-∠ROS=90°......(2)........∠ROQ=90°

From 1 and 2,

∠ROS+∠POS=∠QOS-∠ROS...... transpose ∠ROS on LHS, ∠POS on RHS.

∠ROS+∠ROS=∠QOS-∠POS

2(∠ROS)=∠QOS-∠POS..........transpose 2 to the other side

∠ROS=½(∠QOS-∠POS)

Thus proved!

HOPE THIS HELPS :D

Answer 2

Step-by-step explanation:

POS+ROS+ROQ = 180°

Now, POS+ROS = 180°- 90° (Since POR = ROQ = 90°)

∴ POS + ROS = 90°

Now, QOS = ROQ+ROS

It is given that ROQ = 90°,

∴ QOS = 90° +ROS

Or, QOS – ROS = 90°

As POS + ROS = 90° and QOS – ROS = 90°, we get

POS + ROS = QOS – ROS

2 ROS + POS = QOS

Or, ROS = ½ (QOS – POS) (Hence proved).