WAEC 2018 · Paper 2 · Q2

(a)
In a right-angled triangle, $\sin X = \frac35$ . Evaluate, leaving the answer as a fraction, $5(\cos X)^2 - 3$ .
(b)
The base of a pyramid, $12\text{ cm}$ high, is a rectangle with dimensions $42\text{ cm}$ by $11\text{ cm}$ . If the pyramid is filled with water and emptied into a conical container of equal height and volume, calculate, leaving the answer in surd form, the base radius of the container. $\left[\text{Take }\pi = \frac{22}{7}\right]$

Worked solution (try it first)

$\sin X = \frac35$ gives a right-angled triangle with sides 3, 4 and 5, so $\cos X = \frac45$ .
Then $5(\cos X)^2 - 3 = 5 \times \frac{16}{25} - 3$
$= \frac{16}{5} - 3$
$= \frac15$ .

Volume of the pyramid $= \frac13 \times (42 \times 11) \times 12$
$= 1848\text{ cm}^3$ .
The cone has the same height (12 cm) and volume: $\frac13 \times \frac{22}{7} \times r^2 \times 12 = 1848$ .
Simplify: $\frac{88}{7}r^2 = 1848$ , so $r^2 = 147$ .
In surd form, $r = \sqrt{147}$
$= \sqrt{49 \times 3}$
$= 7\sqrt3$ cm.