Question

A man on the top of a rock lying on a seashore observes a boat coming towards it. If it takes $10$ minutes for the angles of depression to change from ${30}^{\circ }$ to ${60}^{\circ }$ how soon will the boat reach the shore?

• A. $20$ minutes
• B. $15$ minutes
• C. $10$ minutes
• D. $5$ minutes

Answer 1

Answer: (B)

$15$ minutes

AB is the rock C and D be the two positions of the boat

Given:

$\angle EAC={30}^{\circ }$ and $\angle EAD={60}^{\circ }$

Now,

$\angle ADB=EAD$ $[\because AE\parallel BC,AD\text{is transversal},\angle ADB$ and $\angle EAD$ are alternate interior angles $]$
$⟹\angle ADB={60}^{\circ }$

$\angle EAC=ACB$ $[\because AE\parallel BC,AC\text{is transversal},\angle EAC$ and $\angle ACB$ are alternate interior angles $]$

$⟹\angle ACB={30}^{\circ }$

Let $CD=x,DB=y$

From $△ABC$
$\tan {30}^0=\frac{AB}{x+y}=\frac{1}{\sqrt{3}}$
$\Rightarrow AB=\frac{x+y}{\sqrt{3}}$
From $△ABD$

$\tan {60}^0=\frac{AB}{y}\Rightarrow AB=\sqrt{3}y$
$\therefore \frac{x+y}{\sqrt{3}}=\sqrt{3}y$ or $\frac{x}{2}=y$
The boat takes $10$ minutes to cover distance $x,$

so it will take $5$ minutes to cover $\frac{x}{2}$
$\therefore$ Time to reach $B=(10+5)$ min $=15$ Minutes from point $C$