Chapter 4.8, Problem 18E

Solution

To find: The distance and bearing from Fort Lauderdale to the boat.

The distance from the Fort Lauderdale to the boat is approximately 89.44 knots and the bearing is $128.43°$

Given information:

Given, the Cerrito Lindo travels at a speed of 40 knots from Fort Lauderdale on a course of 65° for 2 hr and then changes to a course of 155° for 4 hr.

Calculation:

Considering that the boat travels at same speed, the distance between point A and B is 40 knots and the distance between B and C is 80 knots since the time is twice the initial time.

The triangle representing the data is:

  

Using definitions of trigonometric functions for the triangle ABC:

  $\begin{array}{l}\tan \left(BAC\right)=\frac{\text{opposite side}}{\text{adjacent side}}\\ \tan \left(BAC\right)=\frac{BC}{AB}\\ \angle BAC={\tan }^{-1}\left(\frac{80}{40}\right)\\ \angle BAC\approx 63.43°\end{array}$

Also,

  $\begin{array}{l}A{C}^2=A{B}^2+B{C}^2\\ A{C}^2={\left(40\right)}^2+{\left(80\right)}^2\\ A{C}^2=1600+6400\\ A{C}^2=8000\\ AC=\sqrt{8000}\\ AC\approx 89.44\end{array}$

The distance from the Fort Lauderdale to the boat is approximately 89.44 knots.

The bearing from the Fort Lauderdale to the boat is given by:

  $65°+63.43°=128.43°$

Conclusion:

The distance from the Fort Lauderdale to the boat is approximately 89.44 knots and the bearing is $128.43°$