Question

In the given figure, if POQ is a diameter of the circle and PR=QR, then RPQ is equal to

• A. ${30}^{\circ }$
• B. ${60}^{\circ }$
• C. ${90}^{\circ }$
• D. ${45}^{\circ }$

Answer 1

The correct option is D ${45}^{\circ }$

$\Rightarrow$ In $△PQR$,

$\Rightarrow$ $PR=QR$ [ Given ]

$\therefore$ $\angle RPQ=\angle PQR$ [ Opp angles of equal sides are equal ]

$\Rightarrow$ $\angle QRP={90}^o$ [ Angle inscribed in semi-circle ]

$\Rightarrow$ $\angle PQR+\angle QRP+\angle RPQ={180}^o$

$\Rightarrow$ $\angle RPQ+{90}^o+\angle RPQ={180}^o$

$\therefore$ $2\angle RPQ={180}^o-{90}^o$

$\therefore$ $2\angle RPQ={90}^o$

$\therefore$ $\angle RPQ={45}^o$