A "simple" calc question that i cannot figure out!?

Question

The bearing of a ship from a lighthouse was found to be N 10 degrees E. After the ship sailed 6.6 miles due south, the new bearing was N 30 degrees E. Find the distance between the ship and the light house at each location. Round to the nearest tenth if needed.

Answer 1

As ever, a sketch is indispensable.

Denote the lighthouse's position by L. Let the ship first be at S, and finally at F. Then one has a triangle LSF with known angles and one known side (LS = 6.6m). One can then use the sine formula on this triangle, as follows.

LS /sin 150 = SF / sin 20 ( = 6.6m /sin 20) = LF / sin 10.

So LS = sin 150 * 6.6m /sin 20 = 9.649... m or 9.6 m rounded,

and LF = sin 10 * 6.6m /sin 20 = 3.351... m or 3.3 m rounded.

QED

But tell me, WHY do you call this a "calc" question? It is certainly a "geom./trig." question, but that does not make it a "calc" question. It could however be called a "pre-calc." question.

Live long and prosper.

Answer 2

A drawing would help. Consider yourself at the lighthouse. Consider north to be at 12 o'clock. The first bearing is a line from the lighthouse towards about the position of the hour hand at 12:15. Now the ship turns and heads south. Then the second bearing is a line from the lighthouse towards the position of the hour hand at 1. So you would have a 20 deg, 10 deg, 150 deg triangle with the distance between the 20 degree angle and 150 degree angle of 6.6 miles. Sounds like a law of sines type solution.

Answer 3

They traveled from 10 to 30 at 6.6 miles.
that is a difference of 20 degrees for 6.6 miles

n10 degrees would be 3.3 miles away from the lighthouse and
the distance from ship to the lighthouse at 30 degrees would be 9.9 miles.