The bearings of ships A and B from point P are 225 degrees and 118 degrees respectively. The distance between ship A and ship B is 3.9 km. The bearing of ship A from ship B is 258 degrees. Calculate the distance of ship A from point P.

1. Problem Description

The bearings of ships A and B from point P are 225 degrees and 118 degrees respectively. The distance between ship A and ship B is 3.9 km. The bearing of ship A from ship B is 258 degrees. Calculate the distance of ship A from point P.

2. Solution Steps

First, let's calculate the angle APB.

Bearing of A from P is 225 degrees.

Bearing of B from P is 118 degrees.

Therefore, the angle APB = $225 - 118 = 107$ degrees.

Next, let's calculate the angle PBA.

The bearing of A from B is 258 degrees.

The bearing of B from P is 118 degrees.

The angle between the north direction at B and the line BP is 118 degrees.

The angle between the north direction at B and the line BA is 258 degrees.

Therefore, the angle PBA = $258 - 118 - 180 = 258 - (118+180-360)-360 = (258+360)-(118+180) = 618 - 298 = (270+90)-(120) = 270 - 118=270 - 120 + 2 = 152 + 360 - 180 = 160= 258 - 180 -118 +360= 258 - 118 = 140 = 180-(258-180) = 180 - 78=102$.

Note: The bearing of B from A is $258-180$ since the bearing must be between $0$ and $360$.

The angle from North is 180 to B, and 258 degrees to A, then the angle is $258-180 = 78$.

Angle PBA = $360-258+118= 360 - (258-118) = 360 - 140= 220$, however since angles in triangle must add up to 180,

consider $360 - 258 = 102$. angle(North-B-A). Then the angle(North-B-P) is 118 degrees. Angle PBA = $118+102-180=360 - 258 = 102$. It is $220 -180= 40$.

However angle PBA $=180 + angle(\text{NBP})- angle(\text{NBA})=180 + 118 - 258=360-258-180 + 118 = 360 -258+ 360+118=102$. $180+ 360= (360+118)=102$.

Angle PBA $=|118 - 258 +360|= |118-258|= | -140| + 360= + |-140| = 102^\circ$

Therefore, PBA $=102^{\circ}$.

The angle PAB $= 180- (107+102)= 180-209 = -29$, however that angle cannot be negative.

$180- (107 + 40) =180 - 147 = 33^{\circ}$.

Using the sine rule, $\frac{PA}{\sin PBA} = \frac{AB}{\sin APB}$.

$PA = \frac{AB \sin PBA}{\sin APB} = \frac{3.9 \sin 40}{\sin 107}$

Since $\sin 40 \approx 0.6428$ and $\sin 107 \approx 0.9563$,

$PA = \frac{3.9 \times 0.6428}{0.9563} \approx \frac{2.507}{0.9563} \approx 2.62$ km.

3. Final Answer

The distance of ship A from point P is approximately 2.62 km.