The angle of elevation of the top of a hill from the foot of a tower at B is 50°. The angle of elevation of the top of the tower 100 m high from the foot of the hill at C is 35°.
Find the :
(a) horizontal distance BC between the Hill and the Tower.
(b) height CD of the Hill. (Take tan 50° = 1.20)
(c) time taken by a cyclist to cover the distance BC, cycling at 20 m/sec.
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(a) Given,
Height of tower (AB) = 100 m
∠ACB = 35°
In triangle ABC,
$$\begin{aligned}\Rightarrow \text{tan 35°} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \text{tan 35°} = \frac{AB}{BC} \\[1em] \Rightarrow 0.70 = \frac{100}{BC} \\[1em] \Rightarrow BC = \frac{100}{0.70} \\[1em] \Rightarrow BC = 142.86 m.\end{aligned}$$
Hence, the horizontal distance BC between the Hill and the Tower = 142.86 m.
(b) From part(a),
BC = 142.86 m
From figure,
∠DBC = 50°
In triangle DBC,
$$\begin{aligned}\Rightarrow \text{tan 50°} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \text{tan 50°} = \frac{CD}{BC} \\[1em] \Rightarrow 1.20 = \frac{CD}{142.86} \\[1em] \Rightarrow CD = 142.86 \times 1.20 \\[1em] \Rightarrow CD = 171.43 m.\end{aligned}$$
Hence, height CD of the hill = 171.43 m.
(c) Given,
Speed = 20 m/sec.
By formula,
Time = $\frac{\text{Distance (BC)}}{\text{Speed}}$
= $\frac{142.86}{20}$
= 7.14 seconds.
Hence, time taken by a cyclist to cover the distance BC = 7.14 seconds.