The bearings of ships A and B from a 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 $258^\circ$. Calculate the distance of ship A from point P.

1. Problem Description

The bearings of ships A and B from a point P are $225^\circ$ 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 $258^\circ$. Calculate the distance of ship A from point P.

2. Solution Steps

Let $PA = x$ and $PB = y$.

Bearing of A from P is $225^\circ$. Bearing of B from P is $116^\circ$.

Bearing of A from B is $258^\circ$.

$AB = 3.9$ km.

Angle APB = $225^\circ - 116^\circ = 109^\circ$.

The bearing of B from A is $258^\circ - 180^\circ = 78^\circ$. Thus, the angle between the north direction at A and the line AB is $78^\circ$.

We can determine the angle PAB by subtracting $225^\circ$ from $360^\circ$ and adding $78^\circ$. $360^\circ - 225^\circ = 135^\circ$. So, we have $135^\circ - 78^\circ = 57^\circ$

Also, we can determine the angle PBA by subtracting $116^\circ$ from $258^\circ - 180^\circ = 78^\circ$. This is incorrect.

The bearing of B from A is $258^\circ - 180^\circ = 78^\circ$. Thus $\angle NAB = 78^\circ$, where N is north.

So $\angle PAB = 360 - 225 + (180 - 258) = 135 - 78 = 57^\circ$.

Also $\angle PBA = 258 - 116 -180 = 142-180$. This is impossible

$\angle PBA = 180 - (258-116) = 180 - 142= 38^\circ$.

Then, the angles of the triangle are:

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

$\angle PAB = 57^\circ$.

$\angle PBA = 180^\circ - (258^\circ - 116^\circ) = 180^\circ - 142^\circ = 38^\circ$.

Using the Law of Sines:

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

$\frac{3.9}{\sin(109^\circ)} = \frac{x}{\sin(38^\circ)} = \frac{y}{\sin(57^\circ)}$

$x = \frac{3.9 \times \sin(38^\circ)}{\sin(109^\circ)} = \frac{3.9 \times 0.6157}{0.9455} = \frac{2.401}{0.9455} \approx 2.539$ km.

3. Final Answer

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