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.

(a) Given,

Height of tower (AB) = 100 m

∠ACB = 35°

In triangle ABC,

$\Rightarrow \text{tan 35°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \text{tan 35°} = \dfrac{AB}{BC} \\[1em] \Rightarrow 0.70 = \dfrac{100}{BC} \\[1em] \Rightarrow BC = \dfrac{100}{0.70} \\[1em] \Rightarrow BC = 142.86 m.$

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,

$\Rightarrow \text{tan 50°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \text{tan 50°} = \dfrac{CD}{BC} \\[1em] \Rightarrow 1.20 = \dfrac{CD}{142.86} \\[1em] \Rightarrow CD = 142.86 \times 1.20 \\[1em] \Rightarrow CD = 171.43 m.$

Hence, height CD of the hill = 171.43 m.

(c) Given,

Speed = 20 m/sec.

By formula,

Time = $\dfrac{\text{Distance (BC)}}{\text{Speed}}$

= $\dfrac{142.86}{20}$

= 7.14 seconds.

Hence, time taken by a cyclist to cover the distance BC = 7.14 seconds.