A 6 m tall man finds that the angle of elevation of a 24 m high pillar and the angle of depression of its base are complementary angles. Find the distance of the man from the pillar.

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Solution

The correct option is A
The conditions given in the question can be drawn as below.

$t a n θ = \frac{6}{x} t a n (90 - θ) = c o t θ = \frac{18}{x} S i n c e, t a n θ = \frac{1}{c o t θ} \Rightarrow \frac{x}{18} = \frac{6}{x} \Rightarrow x^{2} = 108 \Rightarrow x = 6 \sqrt{3} m$
Therefore, the distance of the man from the pillar is $6 \sqrt{3} m$

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A 6 m tall man finds that the angle of elevation of a 24 m high pillar and the angle of depression of its base are complementary angles. Find the distance of the man from the pillar.

The correct option is A The conditions given in the question can be drawn as below. tan θ=6xtan(90−θ)=cotθ=18xSince,tanθ=1cotθ⇒x18=6x⇒x2=108⇒x=6√3 m Therefore, the distance of the man from the pillar is 6√3 m