Two workers pull horizontally on a heavy box but one pulls twice as hard as the other. the larger pull is directed at 25.0 west of north and the resultant of these pulls is 430.0n directly northward.

1) Call  F1 the larger force and F1x and F1y its its x-and-y- components.respectively.

I will use the complementary angle: 90 - 25 = 65 to work with the normal convention.

=> cos(65) = F1x / F1 => F1x = - F1*cos(65) (I choose negative as the west direction)

=> sin(65) = F1y / F1 => F1y = F1*sin(65) (I choose positive the north direction)

2) Call F2 the shorter force and F2x and F2y its components

=> cos(x) = F2x / F2 => F2x = F2*cos(x)

=> sin(x) = F2y / F2=> F2y = F2*sin(x)

3) You know that:

- F1 = 2F2
- The net force in the y direction is 430 N
- The net force in the x direction is 0

a)  F1x + F2x = 0

=>  -F1*cos(65) + F2*sin(x) = 0

=> F1*cos(65) = F2 sin(x) => sin(x) = [F1/F2] cos(65)

Remember F1 = 2F2 => F1/F2 = 2 => sin(x) = 2 cos(65) = 0.84524

=> x = arcsin(0.84524) =  57.7

b) F1y + F2y = 430 =>

F1 sin(65) + F2*sin(57.7) = 430 =>

0.9060F1 + 0.84524F2 430

F1 = 2F2 => 0.9060*2F2 + 0.84524F2 = 430 => 1.7512F2 = 430

=> F2 = 430 / 1.7512 = 245.54 N

=> F1 = 2*245.54 =491.1N

There you have the two forces.

The angle of the shorter force is 57.7 measured from the east to the north (this is north of east),  which would be 90 - 57.7 = 32.3 degrees east of north..

 Then the shorter force is 245.5 N at 32.3 degrees east of north

And the larger force is 491.1 N at 25.0 degrees west of north.