a boat sails on a bearing of N15E for 10 km then on a bearing of S85E until it is due east of starting point the distance from the starting point to the nearest km is thena. 10 kmb 38c. 110d. 113e.114
Sketch the situation and you have a triangle ABC, where A is the starting point, B is the turn after 10km, and C is the ending points. From the bearings, angle A is 75° (90° - 15°), angle B is 100° (85° + 15°), and so C must be 5°. Side c (opposite from C) is 10 km. Solve for side b (opposite from B) using the law of sines: b / (sin B) = c / (sin C) b = c * (sin B)/(sin C) = (10 km) * (sin 100°) / (sin 5°) Now, put your calculator in degree mode and see if the answer is closer to 113 or 114.
My answer is a. I plotted the bearings on a plane: starting point is (0,0) and got (15,10), and (85,0). This means the first distance traveled by the boat was the nearest distance to the starting point.
