Thread: A vector problem.

I'm not really sure if vectors are supposed to be posted here, but my course is called calculus and vectors

A ship's course is set at a heading of 143 degrees at 18 knots. A 10-knot current flows at a bearing of 112 degrees. What is the ground velocity of the ship?

Here is a rough diagram I drew:


I translated the 18 knot vector and the faded line is the resultant vector, or the ground velocity that im trying to find. In class we use the cosine law to determine the magnitude, however i'm not really sure how to find the angle opposite of the faded line.. or perhaps i'm going about it wrong?

the upper left angle of the parallelogram is 143 - 112 = 31 degrees

the adjacent angle of the parallelogram (the angle opposite the ground vector) is supplementary to 31 degrees.

Worked out the magnitude of the resultant force to be 27.1 knots. Having a bit of trouble with the angle:

I calculated the angle between the current vector and the resultant vector to be 14.4 degrees. Wouldn't I just add that to 112 degrees to get 126.4? However the book answer has 132 degrees :S

Originally Posted by kmjt

Worked out the magnitude of the resultant force to be 27.1 knots. Having a bit of trouble with the angle:

I calculated the angle between the current vector and the resultant vector to be 14.4 degrees. Wouldn't I just add that to 112 degrees to get 126.4? However the book answer has 132 degrees :S

the angle between the current vector and the resultant is approx 20 degrees and add that to 112 gives 132 .

18/(sin x)=27.1/sin 149