Question : From 40 metres away from the foot of a tower, the angle of elevation of the top of the tower is 60°. What is the height of the tower?
Option 1: $\frac{120}{\sqrt{3}}$ m
Option 2: $\frac{60}{{\sqrt3}}$ m
Option 3: $\frac{50}{{\sqrt3}}$ m
Option 4: $\frac{130}{{\sqrt7}}$ m
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Correct Answer: $\frac{120}{\sqrt{3}}$ m
Solution :
Let height of the tower = $h$ m
From the figure,
$\tan 60°=\frac{h}{40}$
⇒ $\sqrt{3}=\frac{h}{40}$
⇒ $h=40\sqrt{3}$ m
Rationalising we get,
⇒ $ h =40\sqrt{3} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{120}{\sqrt{3}}$ m
Hence, the correct answer is $\frac{120}{\sqrt{3}}$ m.
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