WAEC 2022 · Paper 2 · Q4

Circle geometry

$TU$ is a tangent to the circle $PQRS$ at $P$ . $\angle SPU = 30^\circ$ , $\angle QRS = 70^\circ$ and $|QR| = |RS|$ . Find:

The paper marks this diagram “not drawn to scale”.

(a)
$\angle QPT$ ;
(b)
$\angle PSR$ .

Worked solution (try it first)

(a)

$PQRS$ is a cyclic quadrilateral, so $\angle QPS = 180^\circ - 70^\circ$
$= 110^\circ$ .
$T$ , $P$ , $U$ are on a straight line, so the angles at $P$ add up to $180^\circ$ : $\angle QPT = 180^\circ - 110^\circ - 30^\circ$
$= 40^\circ$ .

(b)

$|QR| = |RS|$ , so triangle $QRS$ is isosceles and $\angle RSQ = \frac{180^\circ - 70^\circ}{2}$
$= 55^\circ$ .
By the alternate segment theorem, the angle between the tangent $PT$ and the chord $PQ$ equals the angle in the alternate segment: $\angle PSQ = \angle QPT = 40^\circ$ .
So $\angle PSR = \angle PSQ + \angle QSR$
$= 40^\circ + 55^\circ$
$= 95^\circ$ .

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