Question

An observer at an altitude of 250 m observes the angle of depression of two boats on the opposite banks of a river to be ${45}^{\circ }$ and ${60}^{\circ }$ respectively. Find the width of the river. Write the answer correct to the nearest whole number. [3 MARKS]

Answer 1

In $\Delta OMA$

$tan{45}^{\circ }=\frac{OM}{AM}$

$I=\frac{250}{x}$ $\{\because tan{45}^{\circ }=1\}$



$\Rightarrow x=250m$

In $\Delta OMB,tan{60}^{\circ }=\frac{250}{y}$

$\Rightarrow \sqrt{3}=\frac{250}{y}\Rightarrow y=\frac{250}{\sqrt{3}}=\frac{250}{1.73}$

$y=144.34$

$\therefore$ Width of river = x + y = 250 + 144.34

= 394.34 m

In $\Delta$ ABC [$\frac{1}{2}$ MARK]

$tan60=\frac{240}{x}$

$\sqrt{3}x=240$

$x=\frac{240}{\sqrt{3}}\dots \dots \left(1\right)$ [1 MARK]

In $\Delta$ ABD

$tan45=\frac{240}{y}$

$1=\frac{240}{y}$

$y=240$ [1 MARK]

Width of river $=240-80\sqrt{3}$

$=80(3-\sqrt{3})$

$=101.44$

$=101m$ [$\frac{1}{2}$ MARK]