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  • In Fig. 10.37, ∠ PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠ OPR.

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

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Construction: Join PS and RS.

PQRS is a cyclic quadrilateral.

So, $\angle \mathrm{PSR}+\angle \mathrm{PQR}=180^{\circ}$
$\Rightarrow \angle P S R+100^{\circ}=180^{\circ}$
$\Rightarrow \angle P S R=180^{\circ}-100^{\circ}=80^{\circ}$
Here, $\angle \mathrm{POR}=2 \angle \mathrm{PSR}$
$\Rightarrow \angle P O R=2 \times 80^{\circ}=160^{\circ}$
In $\triangle \mathrm{OPR}$,
$\mathrm{OP}=\mathrm{OR} \quad$ (Radii)
$\angle \mathrm{ORP}=\angle \mathrm{OPR}$ (the angles opposite to equal sides)

In $\triangle \mathrm{OPR}$,
$\angle \mathrm{OPR}+\angle \mathrm{ORP}+\angle \mathrm{POR}=180^{\circ}$
$\Rightarrow 2 \angle O P R+160^{\circ}=180^{\circ}$ 
$\Rightarrow 2 \angle O P R=180^{\circ}-160^{\circ}$
$\Rightarrow 2 \angle O P R=20^{\circ}$
$\Rightarrow \angle O P R=10^{\circ}$

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