WAEC 2014 · Paper 2 · Q12

(a)
In the diagram, $\overline{TS}$ is a tangent to the circle $PQRS$ at $S$ , and $O$ is the centre of the circle ( $QOS$ is a diameter). If $\angle TSP = 21^\circ$ and $\angle RQP = 100^\circ$ , find, with reasons: (i) $\angle SPR$ ; (ii) $\angle QSR$ .
Separate values with commas, e.g. 3, −2
(b)
Given the relation $T = \sqrt{\dfrac{U}{\frac1f + \frac1g}}$ : (i) make $g$ the subject of the relation; (ii) find $g$ when $T = 3$ , $f = 4$ and $U = 5$ .

Worked solution (try it first)

$\angle PRS = \angle TSP = 21^\circ$ (angle in the alternate segment).
$PQRS$ is a cyclic quadrilateral, so $\angle RSP = 180^\circ - \angle RQP$
$= 180^\circ - 100^\circ$
$= 80^\circ$ (opposite angles).
In $\triangle PRS$ : $\angle SPR = 180^\circ - 80^\circ - 21^\circ$
$= 79^\circ$ (angles in a triangle).

$\angle PQS = \angle PRS = 21^\circ$ (angles in the same segment).
$QS$ is a diameter, so $\angle QPS = 90^\circ$ (angle in a semicircle).
In $\triangle PQS$ : $\angle QSP = 180^\circ - 90^\circ - 21^\circ$
$= 69^\circ$ .
So $\angle QSR = \angle RSP - \angle QSP$
$= 80^\circ - 69^\circ$
$= 11^\circ$ .

Square both sides: $T^2 = \frac{U}{\frac1f + \frac1g}$ .
Combine the fractions underneath: $\frac1f + \frac1g = \frac{g + f}{fg}$ , so $T^2 = \frac{Ufg}{g + f}$ .
Multiply both sides by $(g + f)$ : $T^2 g + T^2 f = Ufg$ .
Collect the $g$ terms on one side: $T^2 f = Ufg - T^2 g = g(Uf - T^2)$ .
So $g = \frac{fT^2}{Uf - T^2}$ .

With $T = 3$ , $f = 4$ , $U = 5$ : $g = \frac{4 \times 9}{5 \times 4 - 9}$
$= \frac{36}{11}$
$= 3\frac{3}{11}$ .