8. Two ships, the Albacore and the Bonito, are $50 \mathrm{~km}$ apart. The Albacore A is $\mathrm{N} 45^{\circ} \mathrm{W}$ of the Bonito. The Albacore sights a distress flare at $S 5^{\circ} \mathrm{E}$. The Bonito sights the distress flare at $\mathrm{S} 50^{\circ} \mathrm{W}$. How far is each ship from the distress flare?

Step 1 ：Let A be the Albacore's position, B be the Bonito's position, and D be the distress flare's position. Let AD be the distance from the Albacore to the distress flare, and BD be the distance from the Bonito to the distress flare.

Step 2 ：We know the angle ABD is 45 degrees, and the distance AB is 50 km. We also know the angles BAD and ADB, which are 5 degrees and 50 degrees, respectively.

Step 3 ：Using the Law of Cosines, we can find the distances AD and BD:

Step 4 ：\[AD = \sqrt{AB^2 + BD^2 - 2(AB)(BD)\cos{\angle ABD}}\]

Step 5 ：\[BD = \sqrt{AB^2 + AD^2 - 2(AB)(AD)\cos{\angle BAD}}\]

Step 6 ：Plugging in the known values, we get:

Step 7 ：\[AD = \sqrt{50^2 + BD^2 - 2(50)(BD)\cos{45^\circ}}\]

Step 8 ：\[BD = \sqrt{50^2 + AD^2 - 2(50)(AD)\cos{5^\circ}}\]

Step 9 ：Solving for AD and BD, we find:

Step 10 ：\[AD \approx 34.20 \mathrm{~km}\]

Step 11 ：\[BD \approx 73.73 \mathrm{~km}\]

Step 12 ：\(\boxed{\text{The Albacore is approximately 34.20 km away from the distress flare, and the Bonito is approximately 73.73 km away from the distress flare.}}\)