A man is watching from the top of a tower, a boat speeding away from the tower. The boat makes an angle of depression of 450with the man's eye when at a distance of 60 meters from the tower. After 5 seconds the angle of depression becomes 300. What is the approximate speed of the boat assuming that it is running in still water?

Given that,

AC = 60 m

∠ ADB = 30^{0}

∠ ACB = 45^{0}

We have the formula,

tanθ = \frac{Height}{Distance}

We have to find the speed of the boat,

AB

is the tower,

C

and

D

are the point of observation from the boat.
Taking the height of tower as h meters,
In

ΔBAC

\tan 45^{0} = \frac{AB}{AC}

\Rightarrow 1 = \frac{AB}{60}

\Rightarrow AB = 60 m

ΔBAD

\tan 30^{o} = \frac{AB}{AD}

\Rightarrow \frac{1}{\sqrt{3}} = \frac{60}{AD}

\Rightarrow AD = 60 \sqrt{3} m

∴ CD = AD - AC

\Rightarrow CD = 60 \sqrt{3} - 60

\Rightarrow CD = 60 (\sqrt{3} - 1)

Then, speed, s

= \frac{60 (\sqrt{3} - 1)}{5} m / s

.
The approximate speed will be,

s = \frac{12 \times 0.73 \times 3600}{1000}

m / s = \frac{3600}{1000} km / h)

\Rightarrow

s=31.536

\Rightarrow s \approx 32 km / h

Thus the approximate speed is 32km/h.
Therefore the correct option is 3.

JEE