The bearings of ships A and B from point P are $225^\circ$ and $116^\circ$ respectively. The distance between ship A and ship B is 3.9 km. The bearing of ship A from ship B is $258^\circ$. We need to calculate the distance of ship A from point P.

1. Problem Description

The bearings of ships A and B from point P are $225^\circ$ and $116^\circ$ respectively. The distance between ship A and ship B is 3.9 km. The bearing of ship A from ship B is $258^\circ$. We need to calculate the distance of ship A from point P.

2. Solution Steps

First, draw a diagram representing the given information. Let $PA = x$. Let $PB = y$. Let $AB = 3.9$. The bearing of A from P is $225^\circ$ and the bearing of B from P is $116^\circ$. The bearing of A from B is $258^\circ$.

The angle $APB = 225^\circ - 116^\circ = 109^\circ$.

To find the angle $PBA$, we need to use the information that the bearing of A from B is $258^\circ$. The bearing of North from B is $0^\circ$. The angle between North and BP is $116^\circ + 180^\circ = 296^\circ$ (subtract 360 if exceeds 360, $296^\circ$ is the bearing of P from B).

The angle $PBA = 296^\circ - 258^\circ = 38^\circ$.

Now we can find angle $PAB$ in triangle $PAB$.

$PAB = 180^\circ - APB - PBA = 180^\circ - 109^\circ - 38^\circ = 33^\circ$.

Using the sine rule, we have:

$\frac{AB}{sin(APB)} = \frac{PA}{sin(PBA)} = \frac{PB}{sin(PAB)}$

$\frac{3.9}{sin(109^\circ)} = \frac{x}{sin(38^\circ)} = \frac{y}{sin(33^\circ)}$

We want to find $x$ (the distance of ship A from point P).

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

$x = \frac{3.9 \cdot 0.6157}{0.9455}$

$x = \frac{2.40123}{0.9455}$

$x \approx 2.5395$

The distance of ship A from point P is approximately 2.54 km.

3. Final Answer

2.54 km