Question

In the figure, if $\angle QPR={67}^0$ and $\angle SPR={72}^0$ and $RP$ is a diameter of the circle, then $\angle QRS$ is equal to:

• A. ${18}^0$
• B. ${23}^0$
• C. ${41}^0$
• D. ${67}^0$

Answer 1

The correct option is C ${41}^0$

Angle in a semicircle is ${90}^0.$
Thus $\angle PQR=\angle PSR={90}^0$
$\angle QPS=\angle QPR+\angle RPS={67}^0+{72}^0={139}^0$
Sum of all angles of quadrilateral must be ${360}^0.$
Thus, ${90}^0+{90}^0+{139}^0+\angle QRS={360}^0$
$\therefore$ $\angle QRS={41}^0$