Question

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

List I		List II
I	$\angle POQ$	P	$28^{\circ }$
II	$\angle OQS$	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: $PS\Vert QR$
I. $\angle OSQ=\angle SQR=28^{\circ }$ $\{$Alternative angles $\}$
In $△OQS$
$OQ=OS\text{ (Radii of same circle) }$
$\Rightarrow \angle OQS=\angle OSQ=28^{\circ }$
(Angles opposite to equal sides are equal)
$\Rightarrow \angle OQS=28^{\circ }$....(i)
II. In $△OQS$
$\angle QOS=180^{\circ }-(\angle OQS+\angle OSQ)$
$=180^{\circ }-\left(28^{\circ }+28^{\circ }\right)\{\text{ from (i) }$
$=180^{\circ }-56^{\circ }=124^{\circ }$
$=\angle OQS=124^{\circ }$
PS is the diameter of the circle.
$\angle POQ+\angle OQS=180^{\circ }$
(By linear pair axiom)
$\Rightarrow \angle POQ=180^{\circ }-\angle OQS$
$=180^{\circ }-124^{\circ }=56^{\circ }$