Question

the angle of elevation of the top of a tower from a point on the ground, which is 15 m away from the foot of the tower is 60°. The height of the tower is what? ​

Answer 1

Question : The angle of elevation of the top of a tower from a point on the ground, which is 15 m away from the foot of the tower is 60°. The height of the tower is ?

Answer :

Height of the tower = 15√3 m


Step - By - Step Explanation :

Let to be required height of tower h Metres

{Refer to attachment }

In ∆ ABC , tan 60° = AB/BC

tan 60° = AB/BC

tan 60° = AB / 15 m

√3 = AB / 15 m

•°• AB = 15√3 m


{ Trigonometry Identity Used }

tan 60 ° = √3

Answer 2

The angle of elevation of the top of a tower from a point on the ground, which is 15 m away from the foot of the tower is 60°. The height of the tower is ?

Refer the attachment for figure.

Solution :

In the given figure, AB is the height of the tower.

BC = 15 m.

Angle of the elevation = theta = /_BCA = 60°

From figure,

$\tt{tan \: \theta = \frac{AB}{ BC}}...(1)$

tan 60° = √3 ... (2)

From (1) and (2) ,

$\tt{\frac{AB}{BC} = 15 \sqrt{ 3}}$

Therefore,

AB = BC × √3

AB = 15 √3 or 26 m.

Considering √3 = 1.73 ( approx. )

Thus, Height of the tower is 15 √3 metres or 26 metres ( approx. ).