Question

In the figure, O is the centre of the circle. $\angle POQ={100}^{\circ };\angle POR=11O^{\circ }$; then $\angle QPR$ equals :

• A. ${210}^{\circ }$
• B. ${200}^{\circ }$
• C. ${150}^{\circ }$
• D. ${75}^{\circ }$

Answer 1

Answer: (D)

${75}^{\circ }$

$\angle QOR+\angle QOP+\angle POR={360}^o$

$\Rightarrow \angle QOR=360-100-110={150}^o$

Angle subtended at the center is twice the angle subtended anywhere on the arc.

$\Rightarrow \angle QPR=\frac{1}{2}\angle QOR={75}^o$

So option D is the right answer.