WAEC 2012 · Paper 2 · Q4

(a)
In the diagram, $\overline{SR}$ is parallel to $\overline{UW}$ , $SVT$ and $UVW$ are straight lines. If $\angle RSP = 45^\circ$ (i.e. $\angle RST = 45^\circ$ ) and $\angle VTU = 20^\circ$ , calculate the values of $x = \angle WVT$ and $y = \angle VUT$ .
Separate values with commas, e.g. 3, −2
(b)
In the diagram, $OPR$ is a right-angled triangle with $\angle ORP = 45^\circ$ , $\angle OQP = 60^\circ$ ( $Q$ on $RP$ ) and $\angle OPR = 90^\circ$ . If $|OR| = 5\text{ m}$ , find, correct to 4 significant figures, $|QP|$ .

Worked solution (try it first)

$SR \parallel UW$ and $SVT$ crosses both, so $\angle WVT = \angle RST = 45^\circ$ (corresponding angles).
So $x = 45^\circ$ .
In $\triangle VUT$ , $\angle WVT$ is an exterior angle, so it equals the sum of the two opposite interior angles: $x = y + 20^\circ$ .
So $y = 45^\circ - 20^\circ = 25^\circ$ .

In $\triangle OPR$ the right angle is at $P$ , so $OR$ is the hypotenuse.
$OP$ is opposite the $45^\circ$ angle at $R$ : $|OP| = 5\sin 45^\circ \approx 3.536$ m.
In $\triangle OPQ$ , also right-angled at $P$ , $OP$ is opposite the $60^\circ$ angle at $Q$ and $QP$ is adjacent: $\tan 60^\circ = \frac{|OP|}{|QP|}$ .
So $|QP| = \frac{3.536}{\tan 60^\circ}$
$= \frac{3.536}{1.732}$
$\approx 2.041\text{ m}$ (4 significant figures).