The bearings of ships A and B from 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 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 need to visualize the positions of ships A, B, and point P. We are given the bearings of A and B from P. Bearing is measured clockwise from the North.

The bearing of A from P is $225^\circ$, and the bearing of B from P is $116^\circ$. This means that angle APB can be calculated using the given information about bearings.

The angle between the North line at P and PA is $225^\circ$. The angle between the North line at P and PB is $116^\circ$. Since $225^\circ > 116^\circ$, the angle APB can be calculated as follows:

The difference in bearings $= 225^\circ - 116^\circ = 109^\circ$.

So, $\angle APB = 109^\circ$.

We are given that the distance between ship A and ship B is $3.9$ km. The bearing of A from B is $258^\circ$. This information can be used to analyze the triangle APB.

We want to find the distance AP.

We can find the angle PAB.

First, let's calculate the angle between North and BA. Since the bearing of A from B is $258^\circ$, the angle between BA and the north direction at B is $258^\circ$. The angle between the north direction at B and BP is $180^\circ + 116^\circ = 296^\circ$ or $116^\circ$. Therefore, $\angle ABP = |258^\circ - (116^\circ+180^\circ)| = |258^\circ - 296^\circ| = |-38^\circ| = 38^\circ$.

Now, we can use the sine rule to find angle PAB.

$\frac{AB}{\sin(\angle APB)} = \frac{AP}{\sin(\angle ABP)} = \frac{BP}{\sin(\angle PAB)}$

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

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

$AP = \frac{3.9 \times 0.6157}{0.9455} \approx \frac{2.40123}{0.9455} \approx 2.5396$

We use the Law of Sines to find PA.

$\frac{AB}{\sin P} = \frac{PA}{\sin B}$

$\frac{3.9}{\sin 109} = \frac{PA}{\sin 38}$

$PA = \frac{3.9\sin 38}{\sin 109}$

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

$PA \approx 2.54$

3. Final Answer

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