An airplane is flying at a height of $300 \; m$ above the ground. Flying at this height, the angles of depression from the airplane of two points on both banks of a river in opposite directions are $45^\circ$ and $60^\circ$ respectively. Find the width of the river. [Given: $\sqrt{3} = 1.732$]
$B_1$, $B_2$: Points on opposite sides of the river bank
$P$: Position of the airplane
$PO$: Height of plane $= 300 \; m$
$B_1B_2$: Width of the river
In $\triangle POB_1$, $\;$ $\dfrac{PO}{OB_1} = \tan 60^\circ = \sqrt{3}$
$\therefore \;$ $OB_1 = \dfrac{PO}{\sqrt{3}} = \dfrac{300}{\sqrt{3}} = 100 \sqrt{3} = 100 \times 1.732 = 173.2 \; m$
In $\triangle POB_2$, $\;$ $\dfrac{PO}{OB_2} = \tan 45^\circ = 1$
$\implies$ $OB_2 = PO = 300 \; m$
$\therefore \;$ Width of river $= OB_1 + OB_2 = 173.2 + 300 = 473.2 \; m$