An observer in a ship sailing on a course of 50 degrees 30 minutes observed a lighthouse with a bearing of N 20 degrees 30 minutes E. after 15 km more sailing on the same course, the bearing of the lighthouse bacame N 39 degrees 30 minutes W. what is the distance of the ship from the lighthouse. I have answered every question in my test paper correctly except for this one..

I guess the question is asking what is the distance AFTER the 15km of additional sailing on the same course.

You can do this by just drawing all the angles relative to North on a piece of paper. However, it's probably a little bit easier to recompute the angles relative to the direction the ship is traveling:

For the first observation, the lighthouse was observed with a bearing of N(20.5)E. Since the ship is traveling on a heading of N(50.5)E, the relative angle is 50.5° - 20.5° = 30°. In other words, if you were standing on the bow of the ship looking straight ahead, the lighthouse would appear 30° to your left.

For the second observation, the relative direction is now 50.5° - (-39.5°) = 90°. I used a negative value for the observed angle since N(39.5)W is the same as N(-39.5)E.

Now you can construct a triangle from the facts known. It will have angles of 30°, 60°, and 90° and one side (the one adjacent to the 30° angle - but not the hypotenuse) of length 15km. Now you can just use SOHCAHTOA to solve for the length of the side opposite of the 30° angle.



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