Let 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 100m. So, AB = 100m.
Speed=DistanceTime
In 2 mins, the man moves from point C to D.
∴ Speed=CD2m/minIn 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=100Also 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