A man in a boat rowing away from a lighthouse 150 m high, takes 2 minutes to change the angle of elevation of the top of the lighthouse from 60° to 45°. Find the speed of the boat.

Let man in the boat be originally at point C and after 2 minutes it reaches the point D and AB be the lighthouse.

AB = 150 meters.

In △ABC,

$\text{tan 60°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \sqrt{3} = \dfrac{AB}{BC} \\[1em] \Rightarrow AB = \sqrt{3} BC \\[1em] \Rightarrow BC = \dfrac{AB}{\sqrt{3}} \\[1em] \Rightarrow BC = \dfrac{150}{\sqrt{3}} \\[1em] \Rightarrow BC = \dfrac{150}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}} \\[1em] \Rightarrow BC = \dfrac{150\sqrt{3}}{3} \\[1em] \Rightarrow BC = 50\sqrt{3} \text{ meters}.$

In △ABD,

$\text{tan 45°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow 1 = \dfrac{AB}{BD} \\[1em] \Rightarrow AB = BD \\[1em] \Rightarrow BD = 150 \text{ metres}.$

CD = BD - BC = 150 - $50\sqrt{3}$

= 150 - 86.6

= 63.4 meters.

In 2 minutes boat covers 63.4 meters or boat covers 63.4 meters in 120 seconds.

Speed = $\dfrac{\text{Distance}}{\text{Time}} = \dfrac{63.4}{120}$ = 0.53 m/sec.

Hence, the speed of boat = 0.53 m/sec.