Question

An observer from the top of lighthouse observes a boat speeding away from the tower. When the boat is at a distance of 100 m from the feet of the tower, the boat makes an angle depression $45°$ with the man's eye. After 10 seconds, the angle of depression becomes $30°$. Assuming the water to be still, calculate the speed of the boat.

• A. $(\sqrt{3}-1)m/sec$
• B. $10(\sqrt{3}-1)m/sec$
• C. $\frac{(\sqrt{3}-1)}{10}m/sec$
• D. $20(\sqrt{3}-1)m/sec$

Answer 1

The correct option is B

$10(\sqrt{3}-1)m/sec$

Step 1: Given,

BC=100m

$\angle ACB=45°$

$\angle ADB=30°$

Let AB= h meter and CD= x meter

Step 2: Formula used,

$\tan \theta =\frac{Perpendicular}{Base}$

First calculate the value of h

$$\begin{gathered} ∆ABC \\ \tan {45}^o=\frac{AB}{BC} \\ \tan {45}^o=\frac{h}{100} \\ 1=\frac{h}{100} \\ h=100m \end{gathered}$$

Step 3: Now calculate the value of x

$$\begin{gathered} ∆ABD \\ \tan {30}^o=\frac{AB}{BD} \\ \frac{1}{\sqrt{3}}=\frac{h}{100+x} \\ 100+x=100\sqrt{3} \\ x=100\left(\sqrt{3}-1\right)m \end{gathered}$$

Step 4: We know the formula,

$speed=\frac{dis\tan ce}{time}$

Let the speed is (v)

$$\begin{gathered} v=\frac{100(\sqrt{3}-1)m}{10sec} \\ v=10(\sqrt{3}-1)m/sec \end{gathered}$$

Hence, option (B) is correct.