Lighthouse B is 8 miles west of lighthouse A. A boat leaves A and sails 5 miles. At this time, it is sighted from B. If the bearing of the boat from B is N62°E, how far from B is the boat?

Answer 1

We have a triangle consisting of lighthouse B, lighthouse A, where the boat started, and C where the boat ends.

BA(referred to as c) = 8, AC (referred to as b) = 5, the direction of BA is 90ºE, and B = 90° - 62° = 28°

We are trying to find BC = a. As we have b, B, and c, we get C with the sin law, and then A by subtracting, then use the cosine law to get A.

sin 28º/5 = sin C/8. C = sin^-1(8 sin 28º/5) = 48.6904834829517° or 180° - 48.6904834829517°.

Then, A = 180° - 28° - 48.6904834829517° = 103.309516517048° or A = 180° - 28° - (180° - 48.6904834829517°) = 48.6904834829517° - 28° = 20.6904834829517°

Then a = 5 sin 103.309516517048°/sin 28° = 10.3642129415667 or 5 sin 20.6904834829517°/sin 28° = 3.76294854417616

The boat can be either 10.3642129415667 miles or 3.76294854417616 miles from lighthouse B.