Question

An observer, 1.7 m tall, is 203–√203 m away from a tower. The angle of elevation from the of observer to the top of tower is 30°. Find the height of tower ?

Answer 1

Let AB be the height of the observer and EC be the height of the tower.

Given:
AB = 1.7 m ⇒ CD = 1.7 m
BC = 203-\[\sqrt{3}\] m 

Let ED be h m.

In ∆ADE,

\[\tan 30^o = \frac{ED}{AD}\]
\[ \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{20\sqrt{3}}\]
\[ \Rightarrow h = 20 m\]

∴ EC = ED + DC = (h + 1.7) m = 21.7 m

Hence, the height of the tower is 21.7 m.