WAEC 2011 · Paper 2 · Q10

(a)
The total surface areas of two spheres are in the ratio $9 : 49$ . If the radius of the smaller sphere is $12\text{ cm}$ , find, correct to the nearest $\text{cm}^3$ , the volume of the bigger sphere. $\left[\text{Take }\pi = \frac{22}{7}\right]$
(b)
A cyclist starts from a point $X$ and rides $3\text{ km}$ due West to a point $Y$ . At $Y$ , he changes direction and rides $5\text{ km}$ North-West to a point $Z$ . (i) How far is he from the starting point, correct to the nearest km? (ii) Find the bearing of $Z$ from $X$ , to the nearest degree.
Separate values with commas, e.g. 3, −2

X at the origin, Y 3 km west, then 5 km north-west to Z.

Worked solution (try it first)

Surface areas are in the ratio of the squares of the radii: $\frac{4\pi(12)^2}{4\pi R^2} = \frac{9}{49}$ .
So $\frac{12^2}{R^2} = \frac{9}{49}$ , and taking square roots, $\frac{12}{R} = \frac37$ .
So $R = 28\text{ cm}$ .
Volume of the bigger sphere $= \frac43\pi R^3$
$= \frac43 \times \frac{22}{7} \times 28^3$
$= \frac43 \times \frac{22}{7} \times 21\,952$
$\approx 91\,989\text{ cm}^3$ .

Draw the diagram.
From $X$ , go 3 km due west to $Y$ .
At $Y$ , draw north.
North-west is the bearing $315^\circ$ , so $YZ$ goes up and to the left, 5 km long.
At $Y$ , the direction back to $X$ is due east ( $090^\circ$ ) and the direction to $Z$ is $315^\circ$ , so the angle inside the triangle is $\angle XYZ = 360^\circ - (315^\circ - 90^\circ)$
$= 135^\circ$ .

Cosine rule: $|XZ|^2 = 3^2 + 5^2 - 2(3)(5)\cos 135^\circ$
$= 34 + 30(0.7071)$
$\approx 55.21$ .
So $|XZ| \approx 7.43$ km, which is 7 km to the nearest km.

Sine rule for the angle at $X$ : $\frac{\sin\angle YXZ}{5} = \frac{\sin 135^\circ}{7.431}$ , so $\sin\angle YXZ = \frac{5 \times 0.7071}{7.431}$
$\approx 0.4758$ and $\angle YXZ \approx 28.4^\circ$ .
At $X$ , $Y$ is due west ( $270^\circ$ ) and $Z$ is $28.4^\circ$ further round clockwise, towards north.
Bearing of $Z$ from $X$ $= 270^\circ + 28.4^\circ$
$\approx 298^\circ$ .