Question

From the top of a building $60m$ and high the angles of depression of the top and the bottom of a tower are observed to be ${30}^{\circ }$ and ${60}^{\circ }.$ Find the height of the tower.

Answer 1

AB$=$EC$=60$m

ED$=$EC$-$DC$=(60-h)m$

In $\Delta$ ADE

$\tan {30}^o=\frac{ED}{AE}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{60-h}{AE}$

$AE=\sqrt{3}(60-h)$

In $\Delta$ AEC

$\tan {60}^o=\frac{EC}{AE}$

$\Rightarrow \sqrt{3}=\frac{60}{AE}$

$AE=\frac{60}{\sqrt{3}}=20\sqrt{3}$

So $\sqrt{3}(60-h)=20\sqrt{3}$

$\Rightarrow 60-20=h$

So $h=40$m.