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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.

Answers (1)

Answer.       [45°]               
Solution.      According to the question.
           
In \bigtriangleupEDC         EC = 20.5 m
DC = 20.5 m
To find 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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