NECO 2024 · Paper 1 · Q34

In the diagram, $PQS$ is a circle with centre $O$ , $RST$ is a tangent at $S$ and $\angle SOP = 120^\circ$ . Find $\angle PST$ .

Worked solution (try it first)

$OS = OP$ (radii), so triangle $OSP$ is isosceles: $\angle OSP = \dfrac{180^\circ - 120^\circ}{2}$
$= 30^\circ$ .
A radius meets a tangent at $90^\circ$ , so $\angle OST = 90^\circ$ .
So $\angle PST = 90^\circ - 30^\circ$
$= 60^\circ$ , option B.