Chapter 4.8, Problem 43E

To determine

To Find: To find the distance from the port.

Answer to Problem 43E

The required angle and distance is $\text{W}23°\text{N}$ and 38.7 nautical miles.

Explanation of Solution

Given information: The speed of the given ship is 10 knots

Formula used: Cosine and Sine formulae are used

Calculation:

Let the port is at point A, ship changes course at point B and the ship be at point C at 4p.m.

Therefore, $\text{AB}=30\text{and}\text{BC}=10$

Using cosine formula,

Let AC $=x$

  $\begin{array}{l}$ \therefore x^2={(30)}^2+{(10)}^2-2\times 30\times 10\cos (180°-34°)$x^2=900+100-600\cos (146°)\\ x^2=1000-600\times -0.83\\ x^2=1000+498\\ x^2=1498\\ x=\sqrt{1498}\\ x=38.7\text{approx}\end{array}$

Thus, distance of ship from port is 38.7 nautical miles.

Also, let the bearing angle be y

Therefore, using sine formula,

  $\begin{array}{l}\frac{\sin A}{BC}=\frac{\sin B}{AC}=\frac{\sin y}{AB}\\ \frac{\sin A}{10}=\frac{\sin (146°)}{38.7}=\frac{\sin y}{30}\\ \frac{\sin A}{10}=0.0144=\frac{\sin y}{30}\\ \frac{\sin y}{30}=0.0144\\ \sin y=0.0144\times 30\\ y=\arcsin (0.4335)\\ y=11°\text{approx}\end{array}$

Hence, the bearing angle is $\text{W}23°\text{N}\text{.}$