Question

From the top of a building 15 m high the angle of elevation of the top of a tower is found to be ${30}^{\circ }$. From the bottom of the same building, the angle of elevation of the top of the tower is found to be ${60}^{\circ }$. Find the height of the tower and the distance between the tower and building.

Answer 1

Suppose AB is the building and CD is the tower.

Let distance between the building and the tower, BC = d m and DE = h m.

In ΔADE,

$tan\theta =\frac{DE}{AE}$

$tan30=\frac{h}{BC}$ (Since AE = BC)

$\frac{1}{\sqrt{3}}=\frac{h}{d}$

$d=\sqrt{3}h$-----(1)

In triangle DBC

$tan60=\frac{DC}{BC}=\frac{DE+EC}{BC}=\frac{DE+AB}{BC}$

$tan60=\frac{h+15}{d}$

$\sqrt{3}=\frac{h+15}{d}$

$d=\frac{h+15}{\sqrt{3}}$ -----(2)

From (1) & (2)

$\sqrt{3}h=\frac{h+15}{\sqrt{3}}$

$h=7.5$

∴ Height of the tower = DC = DE + EC = h + AB = (7.5 + 15) m = 22.5 m

Putting the value of h in (1), we get

$d=12.99$

∴ Distance between tower and building is $12.99$