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Q : 3 In Fig. $\small 10.37$ , $\small \angle PQR=100^{\circ}$ , where P, Q and R are points on a circle with centre O. Find $\small \angle OPR$ .

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Construction: Join PS and RS.

PQRS is a cyclic quadrilateral.

So, $\angle$ PSR + $\angle$ PQR = $180 \degree$

$\Rightarrow \angle PSR+100 \degree=180 \degree$

$\Rightarrow \angle PSR=180 \degree-100 \degree=80 \degree$

Here, $\angle$ POR = 2 $\angle$ PSR

$\Rightarrow \angle POR=2\times 80 \degree=160 \degree$

In $\triangle$ OPR ,

OP=OR (Radii )

$\angle$ ORP = $\angle$ OPR (the angles opposite to equal sides)

In $\triangle$ OPR ,

$\angle$ OPR+ $\angle$ ORP+ $\angle$ POR= $180 \degree$

$\Rightarrow 2\angle OPR+160 \degree= 180 \degree$

$\Rightarrow 2\angle OPR=180 \degree- 160 \degree$

$\Rightarrow 2\angle OPR=20 \degree$

$\Rightarrow \angle OPR=10 \degree$

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