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  • An observer 1.5 metres tall is 20.5 metres away from a tower 22 metres high. Determine the angle of elevation of the top of the tower from the eye of the observer.

An observer 1.5 metres tall is 20.5 metres away from a tower 22 metres high. Determine the angle of elevation of the top of the tower from the eye of the observer.

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best_answer

According to the question.

In \bigtriangleupEDC         EC = 20.5 m
DC = 20.5 m
To find an angle \theta in \bigtriangleupEDC we need to find tan\theta.
\tan \theta = \frac{P}{B}
\tan \theta = \frac{EC}{DC}
\tan \theta = \frac{20\cdot 5}{20\cdot 5}
\tan \theta = 1
\theta = 45^{\circ}                 \left ( \because \tan 45^{\circ}= 1 \right )
Hence the angle of elevation is 45°.

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infoexpert27

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