# Trigonometry

A person on a ship sailing due south at the rate of 15 miles an hour observes a lighthouse due west at 3p.m. At 5p.m. the lighthouse is 52degrees west of north. How far from the lighthouse was the ship at a)3p.m.? b)5p.m.? c)4p.m.?

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1. At 5 pm, the ship is 30 miles south of its 3pm position. The distance from the lighthouse at 3pm is thus

d/30 = tan52°
so, d=38.4

Now, to get the distance at any other elapsed time of t hours,

d^2 = 38.4^2 + (15t)^2

Let 'er rip!

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posted by Steve

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