Chapter 4.8, Problem 17E

Solution

To find: The distance and bearing from the Corpus Christi port to the boat.

The distance from the Corpus Christi port to the boat is approximately 42.43 knots and the bearing is $140°$ .

Given information:

Given, the Coast Guard cutter Angelica travels at 30 knots from its home port of Corpus Christi on a course of 95° for 2 hr and then changes to a course of 185° for 2 hr.

  

Calculation:

Considering that the boat travels at same speed, the distance between point B and C is 30 knots and the distance between B and A is 30 knots as well.

In a triangle when two sides are equal, then the opposite angles are equal i.e., $\angle BCA=\angle CAB=45°$

The triangle representing the data is:

  

Using definitions of trigonometric functions for the triangle ABC:

  $\begin{array}{c}\sin C=\frac{\text{opposite side}}{\text{hypotenuse}}\\ \sin \left(45°\right)=\frac{30}{AC}\\ AC=30\left(\sqrt{2}\right)\\ AC\approx 42.43\end{array}$

The distance from the Corpus Christi port to the boat is approximately 42.43 knots.

The bearing from the Corpus Christi port to the boat is given by:

  $95°+45°=140°$ .

Conclusion:

The distance from the Corpus Christi port to the boat is approximately 42.43 knots and the bearing is $140°$ .