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

6 \sqrt{3} m

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6 m

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2 \sqrt{3} m

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Solution

The correct option is A $6 \sqrt{3} m$

$t a n θ = \frac{6}{x} t a n (90 - θ) = c o t θ = \frac{18}{x} S i n c e, t a n θ = \frac{1}{c o t θ} \Rightarrow \frac{x}{18} = \frac{6}{x} \Rightarrow x^{2} = 108 x = 6 \sqrt{3} m$

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A 6 ft tall man finds that the angle of elevation of a 24 ft 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 6√3m tan θ=6xtan(90−θ)=cotθ=18xSince,tanθ=1cotθ⇒x18=6x⇒x2=108x=6√3m

The correct option is A $6 \sqrt{3} m$

$t a n θ = \frac{6}{x} t a n (90 - θ) = c o t θ = \frac{18}{x} S i n c e, t a n θ = \frac{1}{c o t θ} \Rightarrow \frac{x}{18} = \frac{6}{x} \Rightarrow x^{2} = 108 x = 6 \sqrt{3} m$