Chapter 6, Problem 18RE

To determine

Find the width of the river.

Answer to Problem 18RE

The width of river is approximately $586.32\text{ft}$

Explanation of Solution

Given:

A surveyor finds that a tree on the opposite bank of a river flowing due east has a bearing of $\text{N}{22}^{\circ }{30}^{\circ }\text{E}$ from a certain point and a bearing of N ${15}^{\circ }\text{W}$ from a point 400 feet downstream.

Now sine law can be used to calculate $AC$ and $AB$

If $ABC$ is a triangle with sides $a,b,c,A,B$ and $C$ are the angles opposite to $a,b$ and $c$ respectively then,

  $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$

Therefore, for the triangle ABC above,

  $\frac{400}{\sin 37.5}=\frac{AC}{\sin {67.5}^{\circ }}=\frac{AB}{\sin {75}^{\circ }}$

Take one equality at a time

  $\begin{array}{cc}\frac{400}{\sin 37.5}& =\frac{AC}{\sin {67.5}^{\circ }}\\ AC& =400\left(\frac{\sin {67.5}^{\circ }}{\sin 37.5}\right)\\ & \approx 607\text{ft}\\ \frac{400}{\sin 37.5}& =\frac{AB}{\sin {75}^{\circ }}\\ AB& =400\left(\frac{\sin {75}^{\circ }}{\sin 37.5}\right)\\ & \approx 635\text{ft}\end{array}$

Now in triangle $ACD$ apply sine law,

  $\begin{array}{cc}\frac{AC}{\sin {90}^{\circ }}& =\frac{AD}{\sin {75}^{\circ }}\\ AD& =AC\sin {75}^{\circ }\\ & =607\sin {75}^{\circ }\\ & \approx 586.32\text{ft}\end{array}$

Therefore, the width of river is approximately $586.32\text{ft}$