A man in a boat rowing away from a lighthouse, 150 m high, takes 1.5 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 D and after 1.5 minutes it reaches the point C and AB be the lighthouse.

AB = 150 meters.

In △ABD,

$\Rightarrow \tan 60° = \dfrac{\text{BA}}{AD} \\[1em] \Rightarrow \sqrt3 = \dfrac{150}{a} \\[1em] \Rightarrow a = \dfrac{150}{\sqrt3} \times \dfrac{\sqrt3}{\sqrt3} \\[1em] \Rightarrow a = \dfrac{150\sqrt3}{3} \\[1em] \Rightarrow a = 50\sqrt3 \text{ meters.}$

In △ABC,

$\Rightarrow \tan 45° = \dfrac{BA}{AC} \\[1em] \Rightarrow 1 = \dfrac{150}{a + x} \\[1em] \Rightarrow a + x = 150 \text{ meters.}$

⇒ x = 150 - a

= 150 - 50 $\sqrt3$

= 150 - 86.6

= 63.4 meters.

In 1.5 minutes boat covers 63.4 meters or boat covers 63.4 meters in 90 seconds.

$\text{ speed} = \dfrac{\text{Distance}}{\text{time}} = \dfrac{63.4}{90}$ = 0.70 m/sec.

Hence, the speed of boat = 0.70 m/sec.