At a distance 2hm from the foot of a tower of height hm, the top of the tower and a pole at the top of the tower subtend equal angles. Height of the pole should be
A
5h3m
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B
4h3m
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C
7h5m
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D
3h2m
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Solution
The correct option is A5h3m In ΔABD, tanα=h2h ⇒tanα=12 In ΔABC, tan2α=h+p2h ⇒2tanα1−tan2α=h+p2h ⇒2(12)1−(12)2=h+p2h ⇒43=h+p2h ⇒8h=3h+3p ⇒5h=3p⇒p=5h3m