The bearings of ships A and B from a point P are 225° and 116° respectively. Ship A is 3.9 km from ship B on a bearing of 258°. Calculate the distance of ship A from point P.

1. Problem Description

The bearings of ships A and B from a point P are 225° and 116° respectively. Ship A is 3.9 km from ship B on a bearing of 258°. Calculate the distance of ship A from point P.

2. Solution Steps

First, let's define the locations: P is the point, A and B are the ships. We are given the bearings of A and B from P. The bearing of A from P is 225°, and the bearing of B from P is 116°. The distance from B to A is 3.9 km, and the bearing of A from B is 258°. We want to find the distance PA.

Let's find the angle APB. The bearing of A from P is 225° and the bearing of B from P is 116°. The difference is 225° - 116° = 109°. So, $\angle APB = 109^{\circ}$.

The bearing of A from B is 258°. The bearing of P from B can be calculated as follows:

The bearing of B from P is 116°. Therefore, the bearing of P from B is $116^{\circ} + 180^{\circ} = 296^{\circ}$.

Now, $\angle ABP$ is the difference between the bearing of A from B (258°) and the bearing of P from B (296°). Since 258° < 296°, we have $296^{\circ} - 258^{\circ} = 38^{\circ}$. Hence $\angle ABP = 38^{\circ}$.

In triangle APB, we have $\angle APB = 109^{\circ}$, $\angle ABP = 38^{\circ}$. Thus, $\angle BAP = 180^{\circ} - 109^{\circ} - 38^{\circ} = 180^{\circ} - 147^{\circ} = 33^{\circ}$.

Now we can use the sine rule:

$\frac{PA}{\sin(\angle ABP)} = \frac{BA}{\sin(\angle APB)}$

$\frac{PA}{\sin(38^{\circ})} = \frac{3.9}{\sin(109^{\circ})}$

$PA = \frac{3.9 \times \sin(38^{\circ})}{\sin(109^{\circ})}$

$PA = \frac{3.9 \times 0.6157}{0.9455}$

$PA = \frac{2.401}{0.9455} \approx 2.539$

3. Final Answer

The distance of ship A from point P is approximately 2.539 km.