WAEC 2019 · Paper 2 · Q7

(a)
In the diagram, $MN$ is a chord of a circle with centre $O$ . If $|MN| = 22.42\text{ cm}$ and the perimeter of triangle $MON$ is $55.6\text{ cm}$ , calculate, correct to the nearest degree, $\angle MON$ .
(b)
$T$ is equidistant from $P$ and $Q$ . The bearing of $P$ from $T$ is $060^\circ$ and the bearing of $Q$ from $T$ is $130^\circ$ . (i) Illustrate the information on a diagram. (ii) Find the bearing of $Q$ from $P$ .

Worked solution (try it first)

(a)

$OM$ and $ON$ are radii, so the perimeter is $2r + 22.42 = 55.6$ .
So $2r = 33.18$ and $r = 16.59$ cm.
The perpendicular from $O$ bisects the chord and the angle: $\sin\frac{\angle MON}{2} = \frac{11.21}{16.59}$
$\approx 0.6757$ .
So $\frac{\angle MON}{2} \approx 42.5^\circ$ and $\angle MON \approx 85^\circ$ .

(b)(i)

Draw north at $T$ , with $P$ on $060^\circ$ and $Q$ on $130^\circ$ , the same distance from $T$ .

(ii)

$\angle PTQ = 130^\circ - 60^\circ$
$= 70^\circ$ .
$|TP| = |TQ|$ , so the triangle is isosceles and $\angle TPQ = \frac{180^\circ - 70^\circ}{2}$
$= 55^\circ$ .
At $P$ , the direction back to $T$ is $060^\circ + 180^\circ = 240^\circ$ , and $Q$ is $55^\circ$ further round anticlockwise.
Bearing of $Q$ from $P$ $= 240^\circ - 55^\circ = 185^\circ$ .

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