Physics
posted by Kiley .
Five forces act simultaneously on Point A: the first 60N at 90 degrees, the second 40N at 0 degrees, the third 80N at 270 degrees, the fourth 40N at 180 degrees, and the fifth 50N at 60 degrees. What is the magnitude and direction of a sixth force that produces the equilibriant at Point A?

The sixth force is the opposite (negative) of the sum of the first five vectors. Add them up.
You obviously need some practice in vector addition. 
yeah, i do.
I don't understand how to do it or i would practice. 
I assume you are measuring angle from the +x axis (not North)
Add up the five X components
60 * cos 90 = 0
40 * cos 0 = 40
80 * cos 270 = 0
40 * cos 180 = 40
50 * cos 60 = 25
sum = 25 N in x direction so the sixth has 25 N in x direction
Then y components
60 * sin 90 = 60
40 * sin 0 = 0
80 * sin 270 = 80
40 * sin 180 = 0
50 * sin 60 = 43.3
sum = 23.3 so force six has y component = 23.3
magnitude of F6 = sqrt (25^2 + 23.3^2)
negative x and negative y means quadrant 3
tan theta = 23.3/25
theta = 43 degrees below x so
180+43 = 223 degrees 
tan theta = 23.3/25

25 * ( sin 40 )