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Arun is drawing a picture by standing. His height is 4 ft and length of his image is $4 \sqrt{3} f t$ , then the angle of elevation is

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30°

45°

90°

60°

answer is A.

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Detailed Solution

In $Δ A B C,$ $A B = 4$

$A C = 4 \sqrt{3} f t$

$∠ A C B = θ$

And $\tan θ = \frac{A B}{A C}$

$= \frac{4}{4 \sqrt{3}}$

$\tan θ = \frac{1}{\sqrt{3}}$

$\tan θ = \tan 30 °$

$∴ θ = 30 °$

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