JAMB 1990 · UME · Q38

In a regular pentagon $PQRST$ , $PR$ intersects $QS$ at $O$ . Calculate $\angle RQS$ .

Worked solution (try it first)

Each interior angle of a regular pentagon is $\frac{(5 - 2) \times 180^\circ}{5} = 108^\circ$ , so $\angle QRS = 108^\circ$ .
$QR = RS$ , so triangle $QRS$ is isosceles with equal base angles at $Q$ and $S$ .
$\angle RQS = \frac{180^\circ - 108^\circ}{2}$
$= 36^\circ$ , option A.