WAEC 2011 · Paper 2 · Q4✱✱

(a)
In the diagram, $PQRST$ is a quadrilateral. $PT \parallel QS$ , $\angle PTQ = 42^\circ$ , $\angle TSQ = 38^\circ$ and $\angle QSR = 30^\circ$ . If $\angle QTS = x$ and $\angle PQT = y$ , find: (i) $x$ ; (ii) $y$ .
Separate values with commas, e.g. 3, −2
(b)
In the diagram, $PQRS$ is a circle centre $O$ . If $P\hat OQ = 150^\circ$ , $\angle QSR = 40^\circ$ and $\angle SQP = 45^\circ$ , calculate $\angle RQS$ .

Worked solution (try it first)

(a)(i)

$PT \parallel QS$ , and $TQ$ crosses both, so $\angle TQS = \angle PTQ = 42^\circ$ (alternate angles).
In triangle $QTS$ the angles add up to $180^\circ$ : $x + 42^\circ + 38^\circ = 180^\circ$ , so $x = 100^\circ$ .

(ii)

At $Q$ , the angles along the straight line are $y$ , $\angle TQS = 42^\circ$ and $\angle RQS$ , which the diagram gives as $60^\circ$ .
So $y + 42^\circ + 60^\circ = 180^\circ$ and $y = 78^\circ$ .

(b)

$\angle PSQ$ stands on the same arc $PQ$ as $\angle POQ$ at the centre, so $\angle PSQ = \frac12 \times 150^\circ$
$= 75^\circ$ .
Then $\angle PSR = \angle PSQ + \angle QSR$
$= 75^\circ + 40^\circ$
$= 115^\circ$ .
$PQRS$ is a cyclic quadrilateral, so opposite angles add up to $180^\circ$ : $\angle PQR = 180^\circ - 115^\circ$
$= 65^\circ$ .
So $\angle RQS = \angle PQR - \angle SQP$
$= 65^\circ - 45^\circ$
$= 20^\circ$ .

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