Question

The angle of elevation of the top of a cable tower when observed from the top of a 9 $m$ tall building is 60° and the angle of depression of the foot of this tower is 45°. Determine the height of the cable tower.​​

Answer 1

$AB=DC=9m$

In $\Delta ADC$,
$tan45°=\frac{DC}{AD}$
$\Rightarrow tan45°$ = $\frac{9}{AD}$
$\Rightarrow 1$ = $\frac{9}{AD}$
$\Rightarrow AD=9m$

In $\Delta ADE$,
$tan60°=\frac{ED}{AD}$
$\Rightarrow \sqrt{3}$ $=\frac{ED}{9}$
$\Rightarrow ED$ $=9\sqrt{3}m$

Height of tower = $EC$
$EC=ED+DC$
$\Rightarrow EC=ED+DC$
$\Rightarrow EC=9\sqrt{3}+9=9(\sqrt{3}+1)=24.57m$

So, height of cable tower = 24.57 m