Question

The angle of elevation of the top of a tower at a point on the ground 50 m away from the foot of the tower is 45º. Then the height of the tower (in metres) is
(a)50√3
(b)50
(c)$\frac{50}{\sqrt{2}}$
(d)$\frac{50}{\sqrt{5}}$

Answer 1

Answer:

The height of the tower is 50 m.  

Among the given options option (b) 50 m  is correct.

Step-by-step explanation:

GIVEN:

Distance of a tower from its  foot , BC = 50 m  

Angle of elevation of the top of the tower at a point on the ground, ∠ACB = 45°

Let AB =  'h' m be the height of the tower

In right angle triangle, ∆ABC ,

tan C = AB/BC = P/ H

tan 45° = h/50

1 = h/50

h = 50  

AB = 50 m

Hence , the height of the tower is 50 m.  

HOPE THIS ANSWER WILL HELP YOU…

Answer 2

Answer:

Refer to the attachment for your answer