The problem describes the positions of two ships, A and B, relative to a port P. The bearing of ship A from port P is 225 degrees, and the bearing of ship B from port P is 116 degrees. The distance between ships A and B is 3.9 km, and the bearing of ship A from ship B is 258 degrees. We need to calculate the distance from port P to ship A.

1. Problem Description

The problem describes the positions of two ships, A and B, relative to a port P. The bearing of ship A from port P is 225 degrees, and the bearing of ship B from port P is 116 degrees. The distance between ships A and B is 3.9 km, and the bearing of ship A from ship B is 258 degrees. We need to calculate the distance from port P to ship A.

2. Solution Steps

First, we need to find the angle APB. The bearing of A from P is 225 degrees, and the bearing of B from P is 116 degrees. Therefore, the angle APB is:

$APB = 225 - 116 = 109$ degrees.

Next, we consider the triangle PAB. The distance AB is given as 3.9 km. The bearing of A from B is 258 degrees. The bearing of B from A is 225+180 if 225+180 < 360, or 225+180-360 if 225+180>360 degrees. Since the bearing of A from P is 225 degrees, the bearing of P from A is $225 - 180 = 45$ degrees.

Also, the bearing of B from P is 116 degrees, so the bearing of P from B is $116 + 180 = 296$ degrees.

The angle PBA is calculated from the bearing of A from B (258 degrees) and the bearing of P from B (296 degrees).

So, $PBA = 296 - 258 = 38$ degrees.

Now we have two angles of triangle PAB: APB = 109 degrees and PBA = 38 degrees. We can find the third angle, PAB:

$PAB = 180 - (109 + 38) = 180 - 147 = 33$ degrees.

Now we can use the Law of Sines to find the distance PA.

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

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

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

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

$PA = \frac{2.4012}{0.9455}$

$PA \approx 2.539$ km

3. Final Answer

The distance from port P to ship A is approximately 2.539 km.