JAMB 2012 · UTME · Q27

In the diagram, $PQR$ is a circle with centre $O$ and $PR$ a diameter. If $\angle QPR = x^\circ$ , find $\angle QRP$ .

Worked solution (try it first)

$PR$ is a diameter, so $\angle PQR = 90^\circ$ (angle in a semicircle).
The angles of triangle $PQR$ add up to $180^\circ$ : $\angle QRP = 180^\circ - 90^\circ - x^\circ$ .
So $\angle QRP = (90 - x)^\circ$ , option B.