In Fig. $P O Q$ is a line. Ray $O R$ is perpendicular to line $P Q$ . $O S$ is another ray lying between rays $O P$ and $O R$ . Prove that $∠ R O S = \frac{1}{2} (∠ Q O S - ∠ P O S)$ .

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Solution

$∠ Q O S + ∠ P O S = 180^{\circ} (∵ s u m o f a n g l e s o n a s t r a i g h t l i n e = 180^{\circ}) - (1)$

$A l s o, ∠ P O S + ∠ S O R = 90^{\circ}$

$(∵ ∠ P O R + ∠ R O Q = 180^{\circ}$

$⟹ ∠ P O R + 90 = 180) ⟹ 2 ∠ P O S + 2 ∠ S O R = 180^{\circ} - (2)$

$s u b r a t a c t i n g (1) f r o m (2)$

$⟹ ∠ P O S + 2 ∠ S O R - ∠ Q O S = 0$

$⟹ 2 ∠ S O R = ∠ Q O S - ∠ P O S$

$⟹ ∠ S O R = ∠ R O S = \frac{1}{2} (∠ Q O S - ∠ P O S)$

$H e n c e, p r o v e d .$

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∠ R O S = \frac{1}{2} (∠ Q O S - ∠ P O S) .

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POQ is a line . Ray OR is perpendicular to line PQ . OS is another Ray lying between rays OP and OR . Prove that angle ROS = 1/2(angle QOS - angle POS).

Q. Question 5
In the given figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that

∠ R O S = \frac{1}{2} (∠ Q O S - ∠ P O S) .

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In Fig. POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that ∠ROS=12(∠QOS−∠POS).

∠QOS+∠POS=180∘(∵sumofanglesonastraightline=180∘)−(1)Also,∠POS+∠SOR=90∘(∵∠POR+∠ROQ=180∘⟹∠POR+90=180)⟹2∠POS+2∠SOR=180∘−(2)subratacting(1)from(2)⟹∠POS+2∠SOR−∠QOS=0⟹2∠SOR=∠QOS−∠POS⟹∠SOR=∠ROS=12(∠QOS−∠POS)Hence,proved.

In Fig. $P O Q$ is a line. Ray $O R$ is perpendicular to line $P Q$ . $O S$ is another ray lying between rays $O P$ and $O R$ . Prove that $∠ R O S = \frac{1}{2} (∠ Q O S - ∠ P O S)$ .