Question

Find the angle of elevation of the top of 15° m high tower at a point 15 m away from the base of the tower

Answer 1

AB = 15 m

BC = 15 m


We know that


$= > \tan( \theta ) = \frac{ab}{bc} = \frac{15}{15} = 1 \\ \\ = > \tan( \theta ) = 1 = \tan(45) \\ \\ = > \bf \theta = 45$
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So angle of elevation is 45°

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