WAEC 2016 · Paper 2 · Q3

(a)
In the diagram, $\angle RTS = 28^\circ$ , $\angle VRM = 46^\circ$ and $MQ$ is a tangent to the circle $VRSTU$ at the point $R$ . Find $\angle VUS$ .
(b)
A cylindrical tin, $7\text{ cm}$ high, is closed at one end. If its total surface area is $462\text{ cm}^2$ , calculate its radius. $\left[\text{Take }\pi = \frac{22}{7}\right]$

Worked solution (try it first)

(a)

By the alternate segment theorem, the angle between the tangent $RQ$ and the chord $RS$ equals the angle in the alternate segment: $\angle SRQ = \angle RTS = 28^\circ$ .
Angles on the straight line $MQ$ at $R$ add up to $180^\circ$ : $\angle VRS = 180^\circ - 46^\circ - 28^\circ$
$= 106^\circ$ .
$VUSR$ is a cyclic quadrilateral, so its opposite angles add up to $180^\circ$ : $\angle VUS = 180^\circ - 106^\circ$
$= 74^\circ$ .

(b)

Closed at one end: the surface is the curved side plus one circle, $2\pi rh + \pi r^2 = 462$ .
With $h = 7$ : $\frac{22}{7}(14r + r^2) = 462$ , so $r^2 + 14r = 147$ .
$r^2 + 14r - 147 = 0$ factorises as $(r + 21)(r - 7) = 0$ .
The radius is positive, so $r = 7$ cm.

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