Question

If the angle of elevation of a tower from a distance of 100 metres from its foot is 60°, then the height of the tower is
(a)100√3 m
(b)$\frac{100}{ \sqrt{3} }m$
(c)50√3
(d)$\frac{200}{ \sqrt{3} }m$

Answer 1

Answer:

Among the given options option (a) 100√3 m  is correct.

The height of the tower is 100√3 m.  

Step-by-step explanation:

GIVEN:

Distance of a tower from its  foot , BC = 100 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 tan 60° = h/100

√3 = h/100

h = 100√3

AB = 100√3 m

Hence , the height of the tower is 100√3 m.  

HOPE THIS ANSWER WILL HELP YOU…

Answer 2

Answer:

Refer to the attachment for your answer