Question

The angle of elevation of top of a tower from 2 points at a distance of A and B from the base and in the same straight line with it are complementary. Prove that the height of tower is √AB

PLZ ANSWER CORRECTLY​

Answer 1

EXPLANATION.

Angle of elevation of top of tower from 2 point

at a distance of A and B

To prove.

The height of tower is = √AB

Proof.

$\sf : \implies \: in \: \triangle \: abp \\ \\ \sf : \implies \: \tan( \theta) = \frac{perpendicular}{base} \\ \\ \sf : \implies \: \tan( \theta) = \frac{ab}{bp} = \frac{ab}{a} \: \: \: ....(1)$

$\sf : \implies \: in \: \triangle \: abc \\ \\ \sf : \implies \: \tan(90 - \theta) = \frac{bc}{ab} \\ \\ \sf : \implies \: as \: we \: know \: that \: \tan(90 - \theta) = \cot( \theta) \\ \\ \sf : \implies \: \cot( \theta) = \frac{b}{ab} \: \: .....(2)$

$\sf : \implies \: from \: equation \: (1) \: \: \: and \: \: \: (2) \\ \\ \sf : \implies \: \frac{AB}{a} = \frac{b}{AB} \\ \\ \sf : \implies \: {AB}^{2} = ab \\ \\ \sf : \implies \: AB \: = \sqrt{ab} \\ \\ \sf : \implies \:{hence \: proved}$

Answer 2

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