WAEC 2020 · Paper 2 · Q11

(a)
In the diagram, $MNPQ$ is a circle with centre $O$ , and $MOQ$ is a diameter. $|MN| = |NP|$ and $\angle OMN = 50^\circ$ . Find: (i) $\angle MNP$ ; (ii) $\angle POQ$ .
Separate values with commas, e.g. 3, −2
(b)
Find the equation of the line which has the same gradient as $8y + 4x = 24$ and passes through the point $(-8, 12)$ .

Worked solution (try it first)

(a)(i)

$MNPQ$ is a cyclic quadrilateral, so opposite angles add up to $180^\circ$ : $\angle NPQ = 180^\circ - \angle NMQ$
$= 180^\circ - 50^\circ$
$= 130^\circ$ .
$MQ$ is a diameter, so $\angle MPQ = 90^\circ$ (angle in a semicircle).
So $\angle MPN = 130^\circ - 90^\circ$
$= 40^\circ$ .
$|MN| = |NP|$ , so triangle $MNP$ is isosceles and $\angle NMP = \angle NPM = 40^\circ$ .
So $\angle MNP = 180^\circ - 40^\circ - 40^\circ$
$= 100^\circ$ .

(ii)

$\angle PMQ = \angle NMQ - \angle NMP$
$= 50^\circ - 40^\circ$
$= 10^\circ$ .
The angle at the centre is twice the angle at the circumference on the same arc $PQ$ : $\angle POQ = 2 \times 10^\circ = 20^\circ$ .

(b)

$8y + 4x = 24$ gives $y = -\frac12x + 3$ , so the gradient is $-\frac12$ .
Through $(-8, 12)$ : $y - 12 = -\frac12(x + 8)$ , so $y = -\frac12x - 4 + 12 = -\frac12x + 8$ .

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