Question

For a person standing at a distance of 60 metre from a temple the angle of elevation at its top is 30 °. Find the height of the church


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Answer 1

Answer:-

Given,

AB=60m.

angle A=30°

So,

$\sf \tan(A) = \frac{ AB}{BC }$

$\sf \implies\tan(A) = \frac{60}{BC }$

$\sf \implies\tan(30 \degree) = \frac{60}{BC }$

$\sf \implies \frac{1}{ \sqrt{3} } = \frac{BC}{60}$

$\sf \implies BC = \frac{20\times3}{\sqrt{3}}$

$\sf \implies BC = \frac{20\times \sqrt{3}\times\cancel{\sqrt{3}}}{\cancel{\sqrt{3}}}$

$\sf \implies BC = 20\sqrt{3}$

Hence, the height of the temple is 20√3m.

Answer 2

Answer:

Hence the height of Temple is 20√3m

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