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

1. Problem Description

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

2. Solution Steps

First, we will draw a diagram representing the given information. Let P be the point, and A and B be the positions of the two ships.

The bearing of A from P is $225^\circ$.

The bearing of B from P is $116^\circ$.

The bearing of A from B is $258^\circ$.

The distance AB is 3.9 km. We want to find the distance AP.

We can calculate the angle APB. The bearing of A from P is $225^\circ$ and the bearing of B from P is $116^\circ$. Therefore, the angle APB is $225^\circ - 116^\circ = 109^\circ$.

Now, we need to find the angle PBA. The bearing of A from B is $258^\circ$. The north direction at B is $0^\circ$. The angle between the north direction and the line PB is $116^\circ$. Therefore, the angle between the north direction at B and the line BA is $258^\circ$. So the angle PBA is $258^\circ - 116^\circ - 180^\circ = 62^\circ$. Then $360^\circ - 258^\circ+116^\circ = 218^\circ$. Subtract $180^\circ$ to see that angle PBA is $360-258-180+116 = 38^\circ$

We are given that the bearing of A from B is $258^\circ$. Let $\theta = 258^\circ - 180^\circ = 78^\circ$. The bearing of B from North is $116^\circ$. The angle between North and PB is $116^\circ$. Thus the angle between PB and the east direction is $90^\circ - 116^\circ = -26^\circ$. The angle PBA is $180 - (258 - 180 +180 - 116) = 180-(78+64) = 180 - 142 = 38^\circ$.

The angle PAB can be calculated using the fact that the sum of angles in a triangle is $180^\circ$. So, the angle PAB is $180^\circ - 109^\circ - 38^\circ = 33^\circ$.

Using the sine rule in triangle APB:

$\frac{AP}{\sin(\angle PBA)} = \frac{AB}{\sin(\angle APB)}$

$\frac{AP}{\sin(33^\circ)} = \frac{3.9}{\sin(109^\circ)}$

$AP = \frac{3.9 \times \sin(33^\circ)}{\sin(109^\circ)}$

$AP = \frac{3.9 \times 0.5446}{0.9455}$

$AP = \frac{2.124}{0.9455}$

$AP = 2.246$

3. Final Answer

The distance of ship A from P is approximately 2.25 km.