AP Physics problem

Question

a weight of 5000 N is suspended in equilibrium by two cables. Cable 1 applies a horizontal force to the right of the object and has a tension force of F1. Cable 2 applies a force upward and to the left at an angle of 37 degrees to the negative x-axis and has a tension force F2. what is F2?

Answer 1

Because the upward force on cable 2 must balance the downward 5000 N, and this is the force opposite the angle of the cable, using the fact that sin(angle) is the opposite over hypotenuse, knowing the opposite we want to find the hypotenuse, the tension on cable 2, F2. Thus F2 = 5000/sin(37) = +8308 N. Bonus: The tension on F1 is equal the tension on F2 in the x direction, which is 8308*cos(37) = +6635 N. Verify the math for cable 2: -6635 squared plus 5000 squared = 8308 squared. Thus we have confirmed the tension on F2 is 8308 N.