JAMB 1983 · UME · Q7

Trigonometric ratios

In the figure, $\angle PQR = 60^\circ$ , $\angle QPR = 90^\circ$ , $\angle PRS = 90^\circ$ , $\angle RPS = 45^\circ$ and $QR = 8\text{ cm}$ . Determine $PS$ .

Worked solution (try it first)

In triangle $PQR$ the right angle is at $P$ , so $QR = 8$ cm is the hypotenuse and $PR$ is opposite the $60^\circ$ angle at $Q$ .
So $PR = 8\sin60^\circ$
$= 8 \times \frac{\sqrt3}{2}$
$= 4\sqrt3$ cm.
In triangle $PRS$ the right angle is at $R$ , so $PS$ is the hypotenuse and $PR$ is adjacent to the $45^\circ$ angle: $PS = \dfrac{PR}{\cos45^\circ}$
$= 4\sqrt3 \times \sqrt2$ .
So $PS = 4\sqrt6$ cm, option B.

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