Question

A tower stand vertically on the ground. From a point on the ground 20 m away from the foot of the tower, the angle of elevation of the top of the tower is 60°. What is the height of the tower?

Answer 1

Answer:

The height of the tower is 20√3 m.  

Step-by-step explanation:

GIVEN:

Distance between point on the ground and foot of tower , BC = 20 m  

Angle of elevation of the top of the tower, ∠ACB = 60°

Let AB =  h m be the height of the tower

In right angle triangle, ∆ABC ,

tan C = AB/BC = P/ H

√3 = h/20

h = 20√3

AB = 20√3 m

Hence , the height of the tower is 20√3 m.  

HOPE THIS ANSWER WILL HELP YOU…

Answer 2

Check This Attachment !

_____________

$\large{ \bold{Let \: , }}$ AB be the tower which is standing vertically on the ground and C be the position of the observer

$\therefore$ CA = 20 m

$\angle \:$ACB = 60°

$\large{\fbox{ \fbox{ \bold{To \: Find </p><p> \: \: \: }}}}$

$\to$ Height of the pole = AB = ?

$\fbox{ \fbox{\bold{ \large{ \: Solution \: \: \: }}}}$

$\bold{ \underline{ \: in \: \: Δ ACB \: \: , \: \: \: \: }}$

$\to \bold{ \: Tan \: 60° \: = \frac{AB }{CA }} \\ \\ \to \bold{ Tan \: 60° \: = \frac{ AB }{20 } } \\ \\ \to \bold{ \sqrt{3} = \frac{AB }{20}} \\ \\ \to \bold{ AB = \sqrt{3} \times 20} \\ \\ \to \bold{ AB = 1.732 \times 20} \\ \\ \to \bold{AB = 34.64 \: \: m}$

$\bold{ \underline{Hence \: , \: Height \: of \: the \: pole \: is \: 34.64 \: meter \: \: }}$