WAEC 2021 · Paper 2 · Q12

(a)
In the diagram, $PQRS$ is a circle. $|PQ| = |QS|$ , $\angle SPR = 26^\circ$ and the interior angles of $\triangle PQS$ are in the ratio $2 : 3 : 3$ . Calculate: (i) $\angle PQR$ ; (ii) $\angle RPQ$ ; (iii) $\angle PRQ$ .
Separate values with commas, e.g. 3, −2
(b)
The coordinates of two points $P$ and $Q$ in a plane are $(7, 3)$ and $(5, x)$ respectively, where $x$ is a real number. If $|PQ| = \sqrt{29}$ units, find the values of $x$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

Triangle $PQS$ has angles in the ratio $2 : 3 : 3$ (8 parts of $180^\circ$ ): $45^\circ$ , $67.5^\circ$ and $67.5^\circ$ .
$|PQ| = |QS|$ , so the equal angles are at $P$ and $S$ : $\angle QPS = \angle QSP = 67.5^\circ$ and $\angle PQS = 45^\circ$ .

$\angle SQR$ and $\angle SPR$ stand on the same arc $SR$ , so $\angle SQR = 26^\circ$ (angles in the same segment).
So $\angle PQR = 45^\circ + 26^\circ$
$= 71^\circ$ .

$\angle RPQ = \angle QPS - \angle SPR$
$= 67.5^\circ - 26^\circ$
$= 41.5^\circ$ .

In triangle $PQR$ : $\angle PRQ = 180^\circ - 41.5^\circ - 71^\circ$
$= 67.5^\circ$ .
(Check: $\angle PRQ$ and $\angle PSQ$ stand on the same arc $PQ$ , and $\angle PSQ = 67.5^\circ$ ✓.)

By the distance formula, $|PQ|^2 = (7 - 5)^2 + (3 - x)^2 = 29$ , so $4 + (3 - x)^2 = 29$ and $(3 - x)^2 = 25$ .
So $3 - x = 5$ or $3 - x = -5$ , which gives $x = -2$ or $x = 8$ .