Question

the angle of elevation of the top of a tower from two point at distance of 4cm and x from the foot of the tower is 60° and 30° respectively if so find the height of the tower and the distance x

Answer 1

Answer:

The height of the tower is  $\frac{4\sqrt{3} }{3}$  cm

The distance x is 4/3cm

Step-by-step explanation:

Let the height of the tower be H

∠B = 90°

Therefore, in ΔABC,

tan 30° = $\frac{x}{H}$

=> $\frac{1}{\sqrt{3} }$ = $\frac{x}{H}$

=> H = x$\sqrt{3}$                      ---(i)

In ΔABD,

tan 60° = $\frac{4}{H}$

=> $\sqrt{3}$ = $\frac{4}{H}$

=>  H = 4/ $\sqrt{3}$                   ---(ii)

Equating equation (i) and (ii),

=> x$\sqrt{3}$  = $\frac{4}{\sqrt{3} }$

=> 3x = 4

=> x = 4/3

Therefore, the distance is 4/3cm

Putting the value of x in equation (i),

H =4/3* $\sqrt{3}$    

H= $\frac{4\sqrt{3} }{3}$

Therefore, the height of the tower is  $\frac{4\sqrt{3} }{3}$  cm

Find the attached picture for illustration.