From the top of a cliff, 50 m high, the angle of depression of a buoy is 30°. Calculate to the nearest metre, the distance of the buoy from the foot of the cliff.

Let AB be the cliff.

Let the distance of the buoy from the foot of the cliff be d (BC). Then,

AB (h) = 50 m

From figure,

Angle of Elevation ∠ACB = Angle of Depression ∠CAD = 30° [Alternate interior angles]

In triangle ABC,

We know that,

$\Rightarrow \tan \theta = \dfrac{\text{perpendicular}}{\text{base}} \\[1em] \Rightarrow \tan 30° = \dfrac{AB}{BC} \\[1em] \Rightarrow \dfrac{1}{\sqrt3} = \dfrac{50}{d} \\[1em] \Rightarrow d = 50 \times 1.732 \\[1em] \Rightarrow d = 86.6 \text{ m} .$

Hence, the distance of the buoy from the foot of the cliff 86.6 m.