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Question

# A man rowing a boat away from a lighthouse which is 100 m high takes 2 mins to change the angle of elevation of the lighthouse from 60∘ to 45∘. The speed of the boat is:

A

503 m/min

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B

503(33) m/min

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C

5033 m/min

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D

50313 m/min

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Solution

## The correct option is B 503(3−√3) m/minLet AB be the lighthouse and C, D be the positions of the man when the angle of elevation changes from 60∘ to 45∘. The man has covered a distance CD in 2 min. As per question, the height of the lighthouse is 100 m. So, AB = 100 m. Speed=DistanceTime In 2 mins, the man moves from point C to D. ∴ Speed=CD2m/min In ΔABC, we havetan 60∘=ABBC⇒ √3=100BC⇒BC=100√3×√3√3 ⇒BC=100√33 .... (i) In ΔABD, we have tan 45∘=ABBD⇒1=100BD⇒BD=100 Also CD=DB−BC=100−100√33 [using (i)]=100(1−√33)=100(3−√33)∴ Speed=CD2=100(3−√33)2=50(3−√33)m/min Hence, the required speed of boat is 503(3−√3) m/min

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