Question

Match List-I with List-II.
In the given figure, $P S$ is the diameter of a circle with centre ' $O$ '. If $P S|| Q R$ and $\angle R Q S=28^{\circ}$, match these columns.

List I		List II
I	$\angle P O Q$	P	$28^{\circ}$
II	$\angle O Q S$	Q	$14^{\circ}$
		R	$56^{\circ}$
		S	$40^{\circ}$

• A. I - R; II - S
• B. I - R; II - P
• C. I - Q; II - P
• D. I - Q; II - S

Answer 1

Answer: (B)

Given: $P S \| Q R$
I. $\angle O S Q=\angle S Q R=28^{\circ}$ $\{$Alternative angles $\}$
In $\triangle O Q S$
$ O Q=O S \text { (Radii of same circle) } $
$\Rightarrow \angle O Q S=\angle O S Q=28^{\circ}$
(Angles opposite to equal sides are equal)
$\Rightarrow \angle O Q S=28^{\circ}$....(i)
II. In $\triangle O Q S$
$\angle Q O S =180^{\circ}-(\angle O Q S+\angle O S Q) $
$ =180^{\circ}-\left(28^{\circ}+28^{\circ}\right) \{\text { from (i) }$
$=180^{\circ}-56^{\circ}=124^{\circ}$
$ =\angle O Q S=124^{\circ}$
PS is the diameter of the circle.
$\angle P O Q+\angle O Q S=180^{\circ}$
(By linear pair axiom)
$\Rightarrow \angle P O Q =180^{\circ}-\angle O Q S $
$=180^{\circ}-124^{\circ}=56^{\circ}$