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प्रश्न
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.
विकल्प
30°
45°
60°
90°
MCQ
उत्तर
45°
Explanation:
Let the angle of elevation of the tower from the eye of the observer be `theta`.
Given that:
AB = 22m, PQ = 1.5m = MB
QB = PM = 20.5m
AM = AB - MB = 22 - 1.5 = 20.5m
Now in triangle APM
`"tan" theta = "AM"/"PM"`
`"tan" theta = 20.5/20.5`
`"tan" theta = 1`
`theta = 45^circ`
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