Trigonometry
posted by Ahlelie Reyes .
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.?
Please show the complete solutions and answer. Thank you! :)

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!