A boat is being rowed away from a cliff, 150 m high. At the top of the cliff, the angle of elevation of the boat changes from 60° to 45° in 2 minutes. The speed of the boat is:

1. 1.9 km/hr

2. 2 km/hr

3. 2.4 km/hr

4. 2.5 km/hr

Let height of the cliff be (AB) = 150 m.

Let C and D be the positions of the ships.

In triangle ABC,

$\Rightarrow \tan(60^\circ) = \dfrac{AB}{BC} \\[1em] \Rightarrow \sqrt3 = \dfrac{150}{BC} \\[1em] \Rightarrow BC = \dfrac{150}{\sqrt3} = 50\sqrt3 \text{ m}$

In triangle ABD,

$\Rightarrow \tan(45^\circ) = \dfrac{AB}{BD} \\[1em] \Rightarrow 1 = \dfrac{150}{BD} \\[1em] \Rightarrow BD = 150 \text{ m}.$

Distance the ship moved from position C to D,

CD = BD - BC

CD = 150 - $50\sqrt3$

CD = 50(3 - 1.732)

CD = 50(1.268)

CD = 63.4 m

Distance = 63.4 m = 0.0634 km

Time = 2 min = $\dfrac{1}{30}$ hr

$\Rightarrow \text{Speed} = \dfrac{\text{Distance}}{\text{Time}} \\[1em] = \dfrac{0.0634}{\dfrac{1}{30}} \\[1em] = 0.0634 \times 30 \\[1em] = 1.9 \text{ km/hr}.$

Hence, option 1 is the correct option.