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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 ?
Solution
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,
\[\tan 30^o = \frac{ED}{AD}\]
\[ \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{20\sqrt{3}}\]
\[ \Rightarrow h = 20 m\]
Hence, the height of the tower is 21.7 m.
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