the tension of two cables that are holding a weight.?

Question

a 677 N weight is being held by two cables. The cable on the left is connected to a wall, has a tension of T, and creates an angle of theda with the wall. The cable on the right has a tension of 820 N and makes an angle of 39 degrees with the ceiling. What is the tension of the cable on the left? and what is the angle of theda? PLEASEE HELP!

heres a sloppy picture:

http://i1.tinypic.com/3zqacfc.jpg

Answer 1

Think in terms of forces acting in the x-direction (left-right) and forces acting in the y-direction (up-down). Now, draw the tension forces in the cables as right triangles, so you can find the x and y components. Since the object is not moving, the sum of all forces is equal to 0:
Sum Fx=sum Fy=0

Sum Fx: 820*cos39 - Tsin(t)=0 (T=tension, t=angle theta)
Sum Fy: T*cos(t) + 820*sin39 - 677=0

Fx: 637.3 = Tsin(t), or T = 637.3/sin(t)
Fy: Tcos(t) + 516.0 - 677.0, or T = 161.0/cos(t)

T = 637.3/sin(t) = 161.0/cos(t)
or, 637.3/161.0 = sin(t)/cos(t) = tan(t)

Solve for t and T and you're done.