JAMB 1987 · UME · Q41

A room is 12 m long, 9 m wide and 8 m high. Find the cosine of the angle which a diagonal of the room makes with the floor.

Worked solution (try it first)

The diagonal of the floor: $\sqrt{12^2 + 9^2} = \sqrt{225} = 15$ m.
The room diagonal rises 8 m above that floor diagonal, so it is $\sqrt{15^2 + 8^2} = \sqrt{289} = 17$ m.
The angle with the floor has the floor diagonal as its adjacent side: $\cos\theta = \frac{15}{17}$ , option A.