Question

In the figure given below, $\angle QSR$ is equal to

• A. ${65}^o$
• B. ${50}^o$
• C. ${70}^o$
• D. ${75}^o$

Answer 1

Answer: (A)

Here, $m\angle QPR={50}^{\circ }$

Also $\angle PQO=\angle PRO={90}^{\circ }$ ....... known fact

$\therefore \angle PQO+\angle PRO+\angle QPR+\angle QOR={360}^{\circ }\to$ sum of all Angles

$\therefore {90}^{\circ }+{90}^{\circ }+{50}^{\circ }+\angle QOR={360}^{\circ }$

$m\angle QOR=360-90-90-50$

$\therefore \angle QOR={130}^{\circ }$

Now,

$\angle QSR=\frac{1}{2}\left(\angle QOR\right)\to$ angle substended at center is double by same arc at circumference

$\therefore \angle QSR=\frac{1}{2}\times {130}^{\circ }$

$\therefore m\angle QSR={65}^{\circ }$

$\therefore \angle QSR={65}^{\circ }$