WAEC 2022 · Paper 2 · Q8

Trigonometric ratios
Plane mensuration

In the diagram, $|PR| = 20\text{ cm}$ , $|PS| = 24.12\text{ cm}$ , $\angle PRS = 90^\circ$ and $Q$ is a point on $\overline{PR}$ . If $\angle SPQ = 34^\circ$ and $\angle PSQ = 20^\circ$ , calculate, correct to one decimal place:

(a)
$|SR|$ ;
(b)
$|PQ|$ ;
(c)
the area of $\triangle PQS$ .

Worked solution (try it first)

(a)

Triangle $PRS$ is right-angled at $R$ : $|SR| = \sqrt{24.12^2 - 20^2}$
$= \sqrt{181.77}$
$\approx 13.48$ cm, which is 13.5 cm to one decimal place.

(b)

$\angle RQS$ is an exterior angle of triangle $PQS$ , so $\angle RQS = 34^\circ + 20^\circ$
$= 54^\circ$ .
In right-angled triangle $QRS$ : $|QR| = \frac{13.482}{\tan 54^\circ}$
$\approx 9.796$ cm.
So $|PQ| = 20 - 9.796 \approx 10.2$ cm.

(c)

Triangle $PQS$ has base $PQ$ and height $SR$ (perpendicular to the line $PR$ ): area $= \frac12 \times 10.204 \times 13.482$
$\approx 68.8\text{ cm}^2$ .

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