Question

the angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. if the tower is 50m high. find the height of the building.

Answer 1

I think it may help u!!

Answer 2

Answer:

10 m

Step-by-step explanation:

Refer the attached figure

Height of building = Ab

Height of tower = CD= 50 m

The angle of elevation of the top of a building from the foot of the tower is 30°  i.e. ∠ACB =30°

The angle of elevation of the top of the tower from the foot of the building is 60° i.e. ∠DBC =60°

We are required to find the height of building i.e. AB

In ΔDBC

$tan\theta = \frac{Perpendicular}{Base}$

$tan\theta = \frac{DC}{BC}$

$tan60^{\circ}= \frac{30}{BC}$

$\sqrt{3}= \frac{30}{BC}$

$BC= \frac{30}{\sqrt{3}}$

In ΔABC

$tan\theta = \frac{Perpendicular}{Base}$

$tan\theta = \frac{AB}{BC}$

$tan30^{\circ}= \frac{AB}{ \frac{30}{\sqrt{3}}}$

$\frac{1}{\sqrt{3}}= \frac{AB}{ \frac{30}{\sqrt{3}}}$

$\frac{1}{\sqrt{3}}= \frac{\sqrt{3}AB}{30}$

$\frac{30}{\sqrt{3}\times\sqrt{3} }= AB$

$\frac{30}{3}= AB$

$10= AB$

Thus the height of building is 10 m