## Trigonometry (11th Edition) Clone

Let A be the point where the ship turns on a bearing of $62^{\circ}$ Let B be the point where the ship turns on a bearing of $115^{\circ}$ Let C be the point where the ship turns to the east. The points ABC form a triangle. The angle $A = 90^{\circ}-62^{\circ} = 28^{\circ}$ The angle $B = 62^{\circ}+(180^{\circ}-115^{\circ}) = 127^{\circ}$ The angle $C = 180^{\circ}-28^{\circ}-127^{\circ} = 25^{\circ}$ We can use the law of sines to find the distance $AB$: $\frac{AB}{sin~25^{\circ}} = \frac{50}{sin~127^{\circ}}$ $AB = \frac{50~sin~25^{\circ}}{sin~127^{\circ}}$ $AB = 26.46~miles$ We can use the law of sines to find the distance $BC$: $\frac{BC}{sin~28^{\circ}} = \frac{50}{sin~127^{\circ}}$ $BC = \frac{50~sin~28^{\circ}}{sin~127^{\circ}}$ $BC = 29.39~miles$ The total distance avoiding the icebergs is $26.46+29.39$ which is 55.85 miles. Compared to the original distance of 50 miles, the ship had to travel an extra distance of 5.85 miles.
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