Looks like Steve answered the same question in the first of the "Related Questions" below
http://www.jiskha.com/display.cgi?id=1410269873
from a ship two lighthouses bear north 40 degree east. after the ship sails at 15knots on a course of 135 degrees for 1 hr, 20 min, the lighthouses bear 10 degrees and 345 degrees. find the distance of the ship in the latter position to the farther lighthouse.
3 answers
This one takes some drawing. Let's label the near lighthouse A, and the far one B.
The ship began at P, and ended up at Q.
It helps to draw a line due north from Q. A bit of analysis shows that
PQ = 20 mi
∠BPQ = 95°
∠PBQ = 30°
We want the distance QB, from the later ship's position to lighthouse B.
Using the law of sines,
20/sin30° = BQ/sin95°
BQ = 39.85 naut. mi.
The ship began at P, and ended up at Q.
It helps to draw a line due north from Q. A bit of analysis shows that
PQ = 20 mi
∠BPQ = 95°
∠PBQ = 30°
We want the distance QB, from the later ship's position to lighthouse B.
Using the law of sines,
20/sin30° = BQ/sin95°
BQ = 39.85 naut. mi.
Looks like I used some different angles last time. I hope both methods produce the same answer!!