WAEC 2018 · Paper 2 · Q6

(a)
If $\tan x = \frac{5}{12}$ , $0^\circ < x < 90^\circ$ , evaluate, without using mathematical tables or a calculator, $\dfrac{\sin x}{(\sin x)^2 + \cos x}$ .
(b)
A rectangular lawn measures $14\text{ m}$ by $11\text{ m}$ . A path of uniform width $x\text{ m}$ surrounds it. If the total area of the path is $186\text{ m}^2$ , how wide is the path?

Worked solution (try it first)

(a)

$\tan x = \frac{5}{12}$ gives a right-angled triangle with sides 5, 12 and 13, so $\sin x = \frac{5}{13}$ and $\cos x = \frac{12}{13}$ .
The bottom is $\left(\frac{5}{13}\right)^2 + \frac{12}{13} = \frac{25}{169} + \frac{156}{169}$
$= \frac{181}{169}$ .
So the value is $\frac{5}{13} \div \frac{181}{169} = \frac{5}{13} \times \frac{169}{181}$
$= \frac{65}{181}$ .

(b)

The lawn and path together make a rectangle $(14 + 2x)$ m by $(11 + 2x)$ m (the path adds $x$ on each side).
Area of the path $=$ big rectangle $-$ lawn: $(14 + 2x)(11 + 2x) - 14 \times 11 = 186$ .
Expand: $154 + 50x + 4x^2 - 154 = 186$ , so $4x^2 + 50x - 186 = 0$ , which is $2x^2 + 25x - 93 = 0$ .
Factorise: $(2x + 31)(x - 3) = 0$ .
A width can't be negative, so $x = 3$ m.

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