A boat is being rowed away from a cliff 150 m high. At the top of the cliff the angle of depression of the boat changed from 60° to 45° in 1 minute.
If $\sqrt{3} = 1.73,$ the speed of the boat is
(a) 4.31 km/hr
(b) 3.81 km/hr
(c) 4.63 km/hr
(d) 3.91 km/hr

Open in App

Solution

(b) 3.81 km/h
Let $A B$ be the cliff and $C and D$ be the two positions of the ships. Thus, we have:
$A B = 150$ m, ∠ $A C B = 60^{o}$ and ∠ $A D B = 45^{o}$

Now, in ∆ $A D B$ , we have:
$\frac{A D}{A B} = c o t 45^{o} = 1$

⇒ $\frac{A D}{150} = 1$
⇒ $A D = 150$ m
In ∆ $A C B$ , we have:
$\frac{A C}{A B} = c o t 60^{o}$ $= \frac{1}{\sqrt{3}}$

⇒ $\frac{A C}{150} = \frac{1}{\sqrt{3}}$
⇒ $A C = \frac{150}{\sqrt{3}}$ $= \frac{150 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} = \frac{150 \sqrt{3}}{3} = 50 \sqrt{3}$ m
Now,
$A C = 50 \times 1.73 = 86.5$ m
∴ $C D = (A D - A C)$ $= (150 - 86.5) = 63.5$ m
Speed of the boat = $\frac{Distance}{Time} = \frac{63.5}{60}$ m/s $= (\frac{63.5}{60} \times \frac{18}{5})$ km/h $= 3.81$ km/h

Suggest Corrections

Similar questions

150

^{o}

^{o}

\sqrt{3} = 1.73

60^{o}

45^{o}

150

60^{0}

45^{0}

2.5

50

2

60^{\circ}

45^{\circ}

Join BYJU'S Learning Program