WAEC 2017 · Paper 2 · Q10

(a)
Given that $\sin x = \frac{5}{13}$ , $0^\circ < x < 90^\circ$ , find $\dfrac{\cos x - 2\sin x}{2\tan x}$ .
(b)
A ladder $LA$ leans against a vertical pole at a point $L$ which is $9.6$ metres above the ground. Another ladder $LB$ , 12 metres long, leans on the opposite side of the pole at the same point $L$ . If $A$ and $B$ are 10 metres apart and on the same straight line as the foot of the pole, calculate, correct to two significant figures, the: (i) length of ladder $LA$ ; (ii) angle which $LA$ makes with the ground.
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

$\sin x = \frac{5}{13}$ gives a right-angled triangle with sides 5, 12 and 13, so $\cos x = \frac{12}{13}$ and $\tan x = \frac{5}{12}$ .
Then $\frac{\cos x - 2\sin x}{2\tan x} = \frac{\frac{12}{13} - \frac{10}{13}}{\frac{10}{12}}$
$= \frac{2}{13} \times \frac{12}{10}$
$= \frac{12}{65}$ .

For ladder $LB$ : $|BF| = \sqrt{12^2 - 9.6^2}$
$= \sqrt{51.84}$
$= 7.2$ m.
$A$ and $B$ are on opposite sides of the pole and 10 m apart, so $|AF| = 10 - 7.2 = 2.8$ m.
Then $|LA| = \sqrt{9.6^2 + 2.8^2}$
$= \sqrt{100}$
$= 10\text{ m}$ .

In triangle $LFA$ : $\tan\angle LAF = \frac{9.6}{2.8}$
$\approx 3.429$ , so the angle is about $73.7^\circ$ , which is $74^\circ$ to two significant figures.