Question

19.
The angle of elevation of the top of the tower AB from a point C on the
ground, which is 60 m away from the foot of the tower, is 30°, as shown in
Figure-2. Find the height of the tower.
30°
60 m
Figure-2​

Answer 1

$\huge\mathbb{GIVEN}$

AB is a tower

BC = 60cm

/_ACB = 30°

$\huge\mathbb{SOLUTION}$

In triangle ABC

$\large{tan30°}{=}\frac{AB}{BC}$

1/√3=$\large\frac{AB}{60}$

${60}{=}$$\large\sqrt{3}{AB}$

${AB}{=}$ 3×20/√3

${AB}{=}$${20}\sqrt{3}$m

Answer 2

$\huge\underline\mathrm\red{Given}$

The angle of elevation of the top of the tower AB from a point C on the

ground, which is 60 m away from the foot of the tower, is 30°

$\huge\underline\mathrm\red{To\:find}$

Find the height of tower

$\huge\underline\mathrm\red{Solution}$

According to question

• BC = 60m
• /_ ACB = 30°

In ∆ABC

$\implies\sf tan30\degree=\Large\frac{AB}{BC}$

$\implies\sf \Large\frac{1}{\sqrt{3}}=\Large\frac{AB}{60}$

$\implies\sf 60=\sqrt{3}AB$

$\implies\sf AB=\Large\frac{3×20}{\sqrt{3}}=20\sqrt{3}m$

$\huge\underline\mathrm\red{Note}$

• tan0° = 0
• tan30° = 1/√3
• tan45° = 1
• tan60° = √3
• tan90° = not defined