Two men on either side of a temple of 30 m height observe its top at the angles of elevation 30∘ and 60∘ respectively. Find the distance between the two men. - Sri Chaitanya Infinity Learn Best Online Courses for NCERT Solutions, CBSE, ICSE, JEE, NEET, Olympiad and Class 6 to 12

Let AB be the height of the temple and P and Q are the two men on either side of the temple.

Given that,

$A B = 30 m, ∠ A P B = 30 °, ∠ A Q B = 60 °$

In △ABP,

$\tan (∠ A P B) = \frac{A B }{P B}$

$\tan 30 ° = \frac{30}{P B} $

$P B = 30 \sqrt{3} m$

In △ABP,

$\tan (∠ A Q B) = \frac{A B}{Q B} $

$\tan 60 ° = \frac{30 }{Q B}$

$Q B = \frac{30 }{\sqrt{3}}$

$Q B = 10 \sqrt{3} m$

Now, PQ = PB + QB = $30 \sqrt{3} m + 10 \sqrt{3} m$

So, PQ = $40 \sqrt{3} m$

Therefore, distance between two men, i.e. PQ = $40 \sqrt{3} m$

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