A man in a boat rowing away from a lighthouse $150 \; m$ high takes $1$ minute to change the angle of elevation of the top of the lighthouse from $60^\circ$ to $45^\circ$. Find the speed of the boat in meter per minute. [Given: $\sqrt{3} = 1.732$]
$AB =$ Lighthouse of height $150 \; m$
$C =$ First point of observation
$D =$ Second point of observation
$CD =$ Distance traveled by the boat in $1$ minute
In $\triangle ABD$, $\;$ $\dfrac{AB}{BD} = \tan 45^\circ = 1$
$\implies$ $BD = AB = 150 \; m$
In $\triangle ABC$, $\;$ $\dfrac{AB}{BC} = \tan 60^\circ = \sqrt{3}$
$\implies$ $BC = \dfrac{AB}{\sqrt{3}} = \dfrac{150}{\sqrt{3}} = 50 \sqrt{3} = 86.6 \; m$
Now, $\;$ $CD = BD - BC = 150 - 86.6 = 63.4 \; m$
$\therefore \;$ Distance covered by the boat in $1$ minute $= 63.4 \; m$
$\therefore \;$ Speed of the boat $= 63.4 \; m/\text{min}$