From the base of a pole of height 20 meter, the angle of elevation of the top of a tower i…

Question

From the base of a pole of height 20 meter, the angle of elevation of the top of a tower is $60^\circ$. The pole subtends an angle $30^\circ$ at the top of the tower. Then the height of the tower is:

Accepted Answer

Here AB is a tower and CD is a pole.

In triangle ABC,

$$\tan 60^\circ  = {{AB} \over {AC}} = {{20 + h} \over x}$$ ...... (1)

In triangle BED,

$$\tan 30^\circ  = {h \over x}$$ ...... (2)

Divide equation (1) by equation (2), we get

$${{\tan 60^\circ } \over {\tan 30^\circ }} = {{20 + h} \over x} \times {x \over h}$$

$$ \Rightarrow {{\sqrt 3 } \over {{1 \over {\sqrt 3 }}}} = {{20 + h} \over h}$$

$$ \Rightarrow 3 = {{20 + h} \over h}$$

$$ \Rightarrow 3h = 20 + h$$

$$ \Rightarrow h = 10\,m$$

$$\therefore$$ Height of tower $$ = 20 + 10 = 30\,m$$

Solutions · 1