Use vectors to solve the following problem.
A jet travels at 640 mph 30 degrees west of south for 2 hours. The jet then changes course to 42 degrees north of west. If the jet travels for another 1 1/2 hours on its new course, how far is the jet from its starting point?

Question

Use vectors to solve the following problem. 
A jet travels at 640 mph 30 degrees west of south for 2 hours. The jet then changes course to 42 degrees north of west. If the jet travels for another 1 1/2 hours on its new course, how far is the jet from its starting point?

anonymous · Answer

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anonymous · Answer

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a: distance travel in west of south direction for 2 hours
b: distance travel in north of west direction for 3/2 hours
a = 640 * 2 = 1280 miles
b = 640 * 3/2 = 960 miles

c: distance from starting point
$$c^2 = a^2 + b^2 -2ab \cos (48 + 30) = 1280^2 + 960^2 - 2.1280.960. \cos78$$
Then c^2 = 2560000 - 2457600 cos78
So $$c \approx 1431.45$$ miles