Question

An observer standing at a point $P$ on the top of a hill near the sea-shore notices that the angle of depression of a ship moving towards the hill in a straight line at constant speed is ${30}^o$. After 45 minutes, this angle becomes ${45}^o$. If T (in minutes) is the total time taken by the ship to move to a point in the sea where the angle of depression from $P$ of the ship is ${60}^o$, then T is equal to:

• A. $\frac{45}{\sqrt{3}}-45$
• B. $45(1+\frac{2}{\sqrt{3}})$
• C. $45(2+\frac{1}{\sqrt{3}})$
• D. $45(1+\frac{1}{\sqrt{3}})$

Answer 1

Answer: (D)

$45(1+\frac{1}{\sqrt{3}})$

We know that, time taken to travel between any points is proportional to distance covered.

Hence, $\frac{T_{AC}}{T_{AB}}=\frac{AC}{AB}$ [uniform speed]

In $△PQA$, $\frac{PQ}{AQ}=\tan 30$

$\Rightarrow AQ=\sqrt{3}h$ ....(1)

In $△PBQ$, $\tan 45=\frac{PQ}{BQ}$

$\Rightarrow BQ=h$ ....(2)

Similarly in $△PQC$, $\tan 60=\frac{PQ}{CQ}$

$\Rightarrow CQ=\frac{h}{\sqrt{3}}$

Now $\frac{AC}{AB}=\frac{AQ-CQ}{AQ-BQ}$

$=\frac{\sqrt{3}h-\frac{h}{\sqrt{3}}}{\sqrt{3}h-h}$

$=\frac{\sqrt{3}-\frac{1}{\sqrt{3}}}{\sqrt{3}-1}$

$=1+\frac{1}{\sqrt{3}}$

Thus $T_{AC}=T_{AB}\times (1+\frac{1}{\sqrt{3}})$

$\Rightarrow T=45(1+\frac{1}{\sqrt{3}})$

Hence, the time taken is $45(1+\frac{1}{\sqrt{3}})$.