In the given figure \(\overline {PQ}\) is a line. Ray \(\overline {

1.	In the given figure \(\overline {PQ}\) is a line. Ray \(\overline {OR}\) is perpendicular to line \(\overline {PQ}. \overline {OS}\) os is another ray lying between rays \(\overline {OP}\) and \(\overline {OR}\) Prove that
Answer» Given : OR ⊥ PQ ⇒ ∠ROQ = 90° To prove: ∠ROS = 1/2(∠QOS – ∠POS) From the figure ∠ROS = ∠QOS – ∠QOR ……………(1) ∠ROS = ∠ROP – ∠POS ……………..(2) Adding (1) and (2) ∠ROS + ∠ROS = ∠QOS – ∠QOR +∠ROP – ∠POS [ ∵ ∠QOR = ∠ROP = 90° given] ⇒ 2∠ROS = ∠QOS – ∠POS ⇒ ∠ROS = 1/2 [∠QOS – ∠POS] Hence proved.

In the given figure \(\overline {PQ}\) is a line. Ray \(\overline {OR}\) is perpendicular to line \(\overline {PQ}. \overline {OS}\) os is another ray lying between rays \(\overline {OP}\) and \(\overline {OR}\) Prove that

Answer»

Given : OR ⊥ PQ ⇒ ∠ROQ = 90°

To prove: ∠ROS = 1/2(∠QOS – ∠POS)

From the figure

∠ROS = ∠QOS – ∠QOR ……………(1)

∠ROS = ∠ROP – ∠POS ……………..(2)

Adding (1) and (2)

∠ROS + ∠ROS = ∠QOS – ∠QOR +∠ROP – ∠POS

[ ∵ ∠QOR = ∠ROP = 90° given]

⇒ 2∠ROS = ∠QOS – ∠POS

⇒ ∠ROS = 1/2 [∠QOS – ∠POS]

Hence proved.