WAEC 2014 · Paper 2 · Q3

(a)
In the diagram, $QSP$ and $QTR$ are straight lines, $|PS| = 6\text{ cm}$ , $|QS| = 4\text{ cm}$ , $|QT| = 5\text{ cm}$ and $\angle QTS = \angle RPQ$ . Calculate $|TR|$ .
(b)
If $\sin x = \frac{5}{13}$ , $0^\circ \le x \le 90^\circ$ , evaluate, without tables or a calculator, $\dfrac{\cos x - 2\sin x}{2\tan x}$ .

Worked solution (try it first)

(a)

Triangles $QTS$ and $QPR$ share the angle at $Q$ , and $\angle QTS = \angle QPR$ , so they are similar, with $T$ matching $P$ and $S$ matching $R$ .
Corresponding sides are in the same ratio: $\frac{|QT|}{|QP|} = \frac{|QS|}{|QR|}$ .
Here $|QP| = 4 + 6 = 10$ , so $\frac{5}{10} = \frac{4}{|QR|}$ and $|QR| = 8$ .
So $|TR| = 8 - 5 = 3\text{ cm}$ .

(b)

$\sin x = \frac{5}{13}$ : draw a right-angled triangle with opposite 5 and hypotenuse 13.
The adjacent side is $\sqrt{169 - 25} = 12$ .
So $\cos x = \frac{12}{13}$ and $\tan x = \frac{5}{12}$ .
Then $\frac{\cos x - 2\sin x}{2\tan x} = \frac{\frac{12}{13} - \frac{10}{13}}{\frac{10}{12}}$
$= \frac{2}{13} \times \frac{12}{10}$
$= \frac{12}{65}$ .

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