Question

Find the magnitude of a fourth force on the stone that will make the vector sum of the four forces zero.

Answer 1

100cos30° + 80cos120° + 40cos233° + Dx = 0 
Dx = -22.53 N 

y components: 
100sin30° + 80sin120° + 40sin233° + Dy = 0 
Dy = -87.34 N 

magnitude of D: 
sqrt[(-22.53)² + (-87.34)²] 
90.2 N 

direction of D: 
arctan[(-87.34)/(-22.53)] 
75.5° ref, but since Dx and Dy are both negative, we know this vector is in QIV: 
360 - 75.5° = 284.5°