Question

The angle of elevation of the top of a hill at the foot of a tower is 60° and the angle of elevation of the top of the tower from the foot of the hill is 30°. If the tower is 50 m high, what is the height of the hill?

Answer 1

Height of hill is 150 m

•Height of tower = AB = 50 m

•let height of hill is h m

•Now, In triangle ABC

•tan30° = P/B = AB/BC = 1/√3

• AB/BC = 1/√3

•50/BC = 1/√3

•BC = 50√3 m ______(1)

•Now, In triangle DBC

•tan60° = P/B = DC/BC = √3

•DC/BC = √3

•h/ 50√3 = √3 ( From 1)

•h = 150 m ______(2)

•Height of hill is 150 m

Answer 2

The height of the hill is 150 meters

Step-by-step explanation:

Given as :

A hill and tower standing on same base

The height of tower = h = 50 meters

The height of hill = H meters

The measure of base = x m

The Angle of elevation of the top of a hill at the foot of a tower = 60°

The angle of elevation of the top of the tower from the foot of the hill = 30°

According to question

from figure

In Δ ABC

Tan angle = $\dfrac{perpendicular}{base}$

Or, Tan 60° = $\dfrac{AC}{BC}$

Or, Tan 60° = $\dfrac{H meters}{x meters}$

Or, √3 = $\dfrac{H}{x}$

∴      x = $\dfrac{H}{\sqrt{3} }$  meters                 ............1

Again

In Δ BCD

Tan angle = $\dfrac{perpendicular}{base}$

Or, Tan 30° = $\dfrac{BD}{BC}$

Or, Tan 30° = $\dfrac{h meters}{x meters}$

Or,  $\dfrac{1}{\sqrt{3} }$ = $\dfrac{50}{x}$

∴       x = 30$\sqrt{3}$  meters              ..........2

From eq 1 and eq 2

$\dfrac{H}{\sqrt{3} }$  meters  =   50$\sqrt{3}$  meters    

by cross multiplication

H = 50$\sqrt{3}$ × $\sqrt{3}$

∴  H = 50 × 3

i.e H = 150 meters

So, The height of the hill =   H = 150 meters

Hence, The height of the hill is 150 meters  . Answer