A 331 N stoplight is hanging in equilibrium from cables as shown. The tension in the right cable is 550 N, and it makes an angle of 12 degrees with the horizontal. What is the force of tension exerted by the cable on the left side? 114 N 217 N 538 N 580 N

Let
F=tension of cable on the left
x=angle of left cable makes with the horizontal

Resolve forces vertically (y-direction)
331=550sin(12)+Fsin(x) =>
Fsin(x)=331-550sin(12)=216.649..........(1)

Resolve forces horizontally (x-direction)
Fcos(x)=550cos(12) =>
Fcos(x)=537.981............................(2)

Divide (1) by (2)
tan(x)=Fsin(x)/Fcos(x)=0.40271
x=atan(0.40271) = 21.935 °

From (2),
F=537.981/cos(x)=537.981/0.92761 = 579.965 N

Check:
Fsin(x)+550sin(12)=579.965*sin(21.935)+550*sin(12) = 331.000  ok
Fcos(x)=537.981
550sin(12)=537.981   ok