Question : The angle of elevation of the top of a building at a distance of 70 m from its foot on a horizontal plane is found to be 60°. Find the height of the building.
Option 1: $70 \sqrt{3} \mathrm{~m}$
Option 2: $60 \sqrt{3} \mathrm{~m}$
Option 3: $50 \sqrt{3} \mathrm{~m}$
Option 4: $70 \sqrt{2} \mathrm{~m}$
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Correct Answer: $70 \sqrt{3} \mathrm{~m}$
Solution :
Height = AC
We know, $\tan \theta=\frac{\text{Perpendicular}}{\text{Base}}$
⇒ $\tan60° =\frac{AC}{70}$
⇒ $\sqrt3=\frac{AC}{70}$
⇒ $AC=70\sqrt3\text{ m}$
Hence, the correct answer is $70\sqrt3\text{ m}$.
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