A mountain climber, in the process of crossing between two cliffs by a rope, pauses to rest. She weighs 566 N. As the drawing shows, she is closer to the left cliff than to the right cliff, with the result that the tensions in the left and right sides of the rope are not the same. Find the tensions in the rope to the left and to the right of the mountain climber. (From the figure α = 65.5° and β = 83.0°.) Equations: EFx= 0 since she is at equilibrium EFy= 0 since she is at equilibrium I got up to the part where EFx= Tr Sin 83 -TL Sin 65.5 = 0 and I get Tr = TL Sin 65.5/ Sin 83 EFy = Tr cos 83 + TL cos 65.5 - W = 0 now when you put the two together we get ( TL sin 65.5/ sin 83 ) cos 83 + TL cos 65.5 - w = 0 this is the part where I'm stuck. I don't how they solved for the TL and w together. Do you? Thanks alot!