WAEC 2013 · Paper 2 · Q2

(a)
The diagram shows a triangular prism with $|QR| = |MN| = |OP| = 10\text{ cm}$ and $|NR| = |QM| = 8\text{ cm}$ . If $\angle RON = 90^\circ$ and $\angle RNO = 30^\circ$ , calculate, correct to 3 significant figures, the volume of the prism.
(b)
A bird on top of a tree sights a prey $18\text{ m}$ away and on the same horizontal ground as the foot of the tree. If the height of the tree is $8\text{ m}$ , calculate, correct to the nearest degree, the angle of depression through which the bird sights the prey.

Worked solution (try it first)

The cross-section is the right-angled triangle $RON$ : the hypotenuse is $|NR| = 8\text{ cm}$ and the right angle is at $O$ .
The side opposite the $30^\circ$ angle: $|RO| = 8\sin30^\circ = 4\text{ cm}$ .
The side next to it: $|NO| = 8\cos30^\circ = 4\sqrt3\text{ cm}$ .
Area of the cross-section: $\frac12 \times 4 \times 4\sqrt3 = 8\sqrt3\text{ cm}^2$ .
Volume = area of cross-section × length: $8\sqrt3 \times 10 = 80\sqrt3$
$\approx 138.56$ .
The volume is $139\text{ cm}^3$ to 3 significant figures.

Sketch the tree, 8 m tall, and the prey on the ground 18 m from its foot.
The angle of depression $x$ at the top of the tree equals the angle of elevation at the prey (alternate angles).
In the right-angled triangle, 8 m is opposite $x$ and 18 m is adjacent, so $\tan x = \frac{8}{18}$ .
$x = \tan^{-1}(0.4444) \approx 23.96^\circ$ .
The angle of depression is $24^\circ$ to the nearest degree.