Chapter 9.5, Problem 43PS

Solution

To find: The distance of life guards from windsurfer.

The distance of left and right lifeguards is 194 ft and 213 ft respectively.

Given data: The two life guards L and $L^{\prime }$ at 300 ft distance apart make angles ${40}^{\text{ο}}$ and ${45}^{\text{ο}}$ with windsurfer W as shown in Fig.1.

  

Method/Formula used:

  

In a triangle ABC , sum of angles is ${180}^{\text{ο}}$ and ‘Law of Sines’ states:

  $\angle A+\angle B+\angle C={180}^{\text{ο}}$

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

Calculations:

The two lifeguards are situated at distance $L{L}^{\prime }$ equals 300 ft making angles ${40}^{\text{ο}}$ and ${45}^{\text{ο}}$ respectively. The locations of life guards and the windsurfer form a triangle $\Delta WL{L}^{\prime }$ . Therefore, in $\Delta WL{L}^{\prime }$ ,

  $\begin{array}{c}{40}^{\text{ο}}+{45}^{\text{ο}}+\angle W={180}^{\text{ο}}\\ \angle W={180}^{\text{ο}}-\left({40}^{\text{ο}}+{45}^{\text{ο}}\right)\\ \angle W={95}^{\text{ο}}\end{array}$

Use ‘Law of Sines’ in $\Delta WL{L}^{\prime }$ ,

  $\frac{\sin {95}^{\text{ο}}}{L{L}^{\prime }}=\frac{\sin {40}^{\text{ο}}}{W{L}^{\prime }}=\frac{\sin {45}^{\text{ο}}}{WL}$

Substitute 300 for $L{L}^{\prime }$

  $\frac{\sin {95}^{\text{ο}}}{300}=\frac{\sin {40}^{\text{ο}}}{W{L}^{\prime }}=\frac{\sin {45}^{\text{ο}}}{WL}$ … (1)

From first two ratios of result (1),

  $\begin{array}{c}\frac{\sin {95}^{\text{ο}}}{300}=\frac{\sin {40}^{\text{ο}}}{W{L}^{\prime }}\\ W{L}^{\prime }=\frac{300\sin {40}^{\text{ο}}}{\sin {95}^{\text{ο}}}\end{array}$

Substitute 0.6428 for $\sin {40}^{\text{ο}}$ and 0.9962 for $\sin {95}^{\text{ο}}$

  $\begin{array}{c}W{L}^{\prime }=\frac{300\left(0.6428\right)}{\left(0.9962\right)}\\ =193.58\\ \approx 194\end{array}$

From first and last ratios of result (1),

  $\begin{array}{c}\frac{\sin {95}^{\text{ο}}}{300}=\frac{\sin {45}^{\text{ο}}}{WL}\\ WL=\frac{300\sin {45}^{\text{ο}}}{\sin {95}^{\text{ο}}}\end{array}$

Substitute 0.7071 for $\sin {45}^{\text{ο}}$ and 0.9962 for $\sin {95}^{\text{ο}}$

  $\begin{array}{c}W{L}^{\prime }=\frac{300\left(0.7071\right)}{\left(0.9962\right)}\\ W{L}^{\prime }=212.94\\ \approx 213\end{array}$

Thus, the distance of left and right lifeguards is 194 ft and 213 ft respectively.