A moving boat is observed from the top of a 150 m high cliff moving away from the cliff. The angle of depression of the boat changes from to

A moving boat is observed from the top of a 150 m high cliff moving away from the cliff. The angle of depression of the boat changes from $60^{\circ}$ to $45^{\circ}$ in 2 minutes. Find the speed of the boat in m/min.

Answers (1)

Given the height of diff, h = 150 m
the initial angle of depression = $60^{\circ}$
                final angle of depression = $45^{\circ}$
Time taken by boat $= 2\, minutes$
To find $\rightarrow$ Speed of boat, $\vartheta = ?$
Look at the figure to get a clear understanding

Boat was initialy at C and finally at D
In $\triangle ABC$
$\tan 60^{\circ}= \frac{AB}{BC}= \frac{h}{BC}= \frac{150}{BC}$
$\Rightarrow BC= \frac{150^{\circ}}{\tan 60^{\circ}}= \frac{150}{\sqrt{3}}m---(i)$
In $\triangle ABD$
$\tan 45^{\circ}= \frac{AB}{BD}= \frac{h}{BD}= \frac{150}{BD}$
$\Rightarrow BD= \frac{150}{\tan 45^{\circ}}= \frac{150}{1}= 150m---(ii)$
$\Rightarrow CD= BD-BC$
$CD=150-\frac{150}{\sqrt{3}}$ (from (i) and (ii))
$CD=150\left ( 1-\frac{1}{\sqrt{3}}\right )m= 150\frac{\left ( \sqrt{3}-1 \right )}{\sqrt{3}}m$
Speed of boat in m/min $= \frac{d}{t}= \frac{CD}{2}= \frac{150\frac{\left ( \sqrt{3}-1 \right )}{\sqrt{3}}}{2}$
                              $\vartheta = 75 \frac{\left ( \sqrt{3}-1 \right )}{\sqrt{3}}\, m/min$
                             $\vartheta = 75 \frac{\left ( 1\cdot 73-1 \right )}{1\cdot 73}$
                            $\vartheta = 75\times \frac{0\cdot 73}{1\cdot 73}= 31\cdot 64\, m/min$