The bearings of ships A and B from port 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 port P.

1. Problem Description

The bearings of ships A and B from port 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 port P.

2. Solution Steps

Let A, B, and P represent the positions of ship A, ship B, and port P, respectively. We are given the bearings of A and B from P, so we have the angles $\angle NPA = 225^{\circ}$ and $\angle NPB = 116^{\circ}$, where N is the north direction from P. Thus, the angle $\angle APB$ within the triangle APB can be determined.

We are given that the bearing of A from B is $258^{\circ}$. The angle $\angle NBA = 258^{\circ}$.

We have $\angle APB = |\angle NPA - \angle NPB| = |225^{\circ} - 116^{\circ}| = 109^{\circ}$. However, this is the exterior angle at $P$. The interior angle $\angle APB$ inside $\triangle APB$ is $109^\circ$, not the external angle.

Since the bearing of A from B is $258^{\circ}$, the angle between North and BA is $258^{\circ}$. Then the angle between the south and BA is $258^{\circ} - 180^{\circ} = 78^{\circ}$. The angle between north and BP is $116^{\circ}$. Therefore, the angle $\angle ABP = 360^{\circ} - 258^{\circ} + 180^{\circ} - 116^{\circ} = 84 + 64 = 86^{\circ} - (360^{\circ} - 258^{\circ}) = 180^{\circ} - (258^{\circ} - 180^{\circ}) = 180^{\circ} - 78^{\circ} = 102^{\circ}$. We are looking for the internal angle at B, which is $360^{\circ} - (258^{\circ} + 180^\circ - 116^{\circ}) = 180^{\circ} - (258^{\circ} - 180^{\circ}) - (116^\circ - 90^\circ)=360^{\circ} - (258 - 180) = 102$. No, the angle APB is between $109$.

The angle at B inside the triangle is $360 - (258 + 180 - 116) = 360 - 322 = 38^{\circ}$.

Since we know two angles, $\angle APB = 109^{\circ}$ and $\angle ABP = 38^{\circ}$, we can find the third angle:

$\angle PAB = 180^{\circ} - \angle APB - \angle ABP = 180^{\circ} - 109^{\circ} - 38^{\circ} = 33^{\circ}$.

We are given that AB = 3.9 km. We want to find the length AP.

Using the Sine Rule:

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

$AP = \frac{AB \cdot \sin(\angle ABP)}{\sin(\angle APB)} = \frac{3.9 \cdot \sin(38^{\circ})}{\sin(109^{\circ})} = \frac{3.9 \cdot 0.6157}{0.9455} \approx \frac{2.401}{0.9455} \approx 2.539$

3. Final Answer

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