Direction problem; find angle to shortest path with wind

[caveman127]

Homework Statement

A pilot must travel directly to a town 500 miles away in a directions 20 degrees east of south from her present position. There is a steady 40 mph wind blowing west to east. Her plane's cruising speed through calm air is 95 mph. Using vector, what must be the plane's heading.

Homework Equations

tan^-1(A_y/A_x) = angle of vector A.

The Attempt at a Solution

V_p/e = 95 mph south
V_a/e = 40 mph east
Add vectors to get V_p/a = -95i + 40j → |V_p/a| = 103.078 mph
Angle of V_p/a = tan^-1(-95/40) = -67.2 degrees from...east? so 67.2 south of east.

I feel like this is pretty far off, because I do anything with the fact the town is 500 miles away and 20 degrees east from south. Please help if you can :)

 

[tiny-tim]

welcome to pf!

hi caveman127! welcome to pf!

V_p/e = 95 mph south

no, V_p/a = 95 mph

(and where did you get south from?)

 

[barryj]

You might start by drawing a somewhat scaled vector diagram of this situation and you will see that 67.2 S of East is wrong. Also, I think that 103.078 is incorrect also. Once you get a good diagam, this is a trig problem.