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प्रश्न
An observer, 1.5m tall, is 28.5m away from a tower 30m high. Determine the angle of elevation of the top of the tower from his eye.
उत्तर
Here, ED is the height of the observer and AC is the tower.
BE = CD = 28. 5 m
AB = AC - BC = 30 m - 1.5 m = 28.5 m
In ΔABE,
`tan <"ABE" = "AB"/"BE"`
⇒ `tan <"ABE" = (28.5"m")/(28.5"m") = 1`
But, `tan 45^circ = 1`
`therefore <"ABE" = 45^circ`
Thus , the required angle of elevation is `45^circ`,
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