1. Feb 1, 2005

### Mivz18

I'm having trouble with this one problem:

A plane flies from base camp to lake A, 280 km away, in a direction of 20.0 degrees north of east. After dropping off supplies it flies to lake B, which is 190 km at 30.0 degrees west of north from lake A. Graphically determine the distance and direction from lake B to the base camp.

Well, I began by drawing the two paths making an angle of 30 facing west. Drawing the third path to make a triangle, I then find the vector x and y components. After I get all my numbers and finish calculating, I get an answer of close to 400 km and 50 degrees south of west. However, the book gives an answer of 310 km at 57 degrees south of west. Any guidance or suggestions as to what to do?

2. Feb 1, 2005

### HallsofIvy

"Two paths making an angle of 30 facing west"? What does that mean?

Since the first leg is "north of east" and the second leg is "west of north" you certainly won't wind up SOUTH of west.
It is standard to take north "up" and east "to the right". If you do that then the first leg can be represented by a line of length 280 at 20 degrees above the horizontal . The second leg will be a line, starting at "lake A", of length 190, 30 degrees to the left of the vertical (and so 90-30= 60 degrees above the horizontal).

The first leg is (by "alternate interior angles") 20 degrees below that same horizontal.
The angle between the two legs is 60+ 20= 80 degrees.

Drawing the third leg to form a triangle gives you a triangle with two sides of length 280 and 190, with an 80 degree angle between them. You can use the "cosine law" to find the length of the third side and the sine law to determine the angles.