Two ships leaves port at the same time one travels 5km/h on a bearing of 045° the other travels at 9km/h on a bearing of 127°. How far apart are the ships after 2 hours?
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Draw the vector triangle with
A = 5 * 2 = 10 km
B = 9 * 2 = 18 km
The third side of this triangle C is the difference between the 2 vectors
Using the Law of Cosines A^2 + B^2 - 2 A B cos theta = C^2
10^2 + 18^2 - 2 * 10 * 18 cos 82 = 374
C = 19.3 km
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