Question

The angle of depression from the top of a tower to the base of an electric-pole is ${60}^{\circ }$ and to the top of the pole is ${30}^{\circ }$. If height of the tower is $15$m,
then what is the height of the electric pole?

• A. $5$ m
• B. $8$ m
• C. $10$ m
• D. $12$ m

Answer 1

The correct option is C $10$ m

Given that, Height of tower = $15$ m.
In $\Delta$ ABC,
tan $\theta =\frac{AB}{BC}$
$tan{60}^{\circ }=\frac{15}{BC}$
$\sqrt{3}=\frac{15}{BC}$
BC = $\frac{15}{\sqrt{3}}=\frac{15\sqrt{3}}{3}=5\sqrt{3}$m.
In $\Delta$ ADE,
tan $\theta =\frac{AE}{DE}$
$tan{30}^{\circ }=\frac{AE}{5\sqrt{3}}$ [$\therefore BC=DE=5\sqrt{3}$]
$\frac{1}{\sqrt{3}}=\frac{AE}{5\sqrt{3}}$
AE = $5m$.
Height of electric pole = $15-5=10$ m.