WAEC 2016 · Paper 2 · Q3

(a)
If $x = -1$ , $y = -3$ , $z = -4$ and $w = -7$ , evaluate $\dfrac{x^3 - y^2}{2w - z}$ .
(b)
In the diagram, $\overline{MN}$ and $\overline{MQ}$ are tangents to the circle centre $O$ , and $PQ$ is a diameter. If $\angle MNQ = x$ , $\angle NMQ = y$ and $\angle NQP = 46^\circ$ , find the value of: (i) $x$ ; (ii) $y$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

Substitute, with brackets round each negative number: $\frac{(-1)^3 - (-3)^2}{2(-7) - (-4)} = \frac{-1 - 9}{-14 + 4}$
$= \frac{-10}{-10}$
$= 1$ .

(b)(i)

The tangent $MQ$ meets the radius $OQ$ at $90^\circ$ , and $PQ$ is a diameter, so $\angle MQP = 90^\circ$ .
Then $\angle MQN = 90^\circ - 46^\circ$
$= 44^\circ$ .
Tangents from $M$ are equal ( $MN = MQ$ ), so triangle $MNQ$ is isosceles and $x = \angle MNQ = \angle MQN = 44^\circ$ .

(ii)

In triangle $MNQ$ : $y = 180^\circ - 44^\circ - 44^\circ$
$= 92^\circ$ .

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