Question

From the top of a building, the angle of elevation of the top of a cable tower is 60º and the angle of depression of its foot is 30º. If the height of the building is 30 m, then the height of the cable tower is

• A. 90 m
• B. 95 m
• C. $30\sqrt{3}m$
• D. 120 m

Answer 1

The correct option is D 120 m

Let AB be the cable tower and CD be the building of height 30 m.
Let AB = h.
Clearly, BE = CD = 30 m
And, AE = AB – BE
⇒ AE = h – 30 …..(i)
$In\Delta DCB,$
$\tan {30}^{\circ }=\frac{DC}{BC}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{30}{BC}$
$\Rightarrow BC=30\sqrt{3}m$
$\Rightarrow BC=ED=30\sqrt{3}m$ ....(ii)
$In\Delta AED,$
$\tan {60}^{\circ }=\frac{AE}{ED}$
$\Rightarrow \sqrt{3}=\frac{h-30}{30\sqrt{3}}$ [From (i) and (ii)]
$\Rightarrow 90=h-30$
$\Rightarrow h=120m$
Thus, the height of the cable tower is 120 m.
Hence, the correct answer is option (d).