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.
