Rope tension on a mountain climber

[Firestrider]

Homework Statement

A mountain climber, in the process of crossing between two cliffs by a rope, pauses to rest. She weighs 586 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 = 60.0° and β = 88.0°.)

Homework Equations

N/A

The Attempt at a Solution

-T_L*cos(60) + T_R*cos(88) = 0
T_L*sin(60) + T_R*sin(88) = 586N

T_L*cos(60) = T_R*cos(88)
T_L = T_R*cos(88) / cos(60)
T_L = 0.07*T_R

0.07*T_R*sin(60) + T_R*sin(88) = 586N
T_R = 586N / (0.07*sin(60) + sin(88)) <-- This might be where I'm confused
T_R = 552.31N
T_L = 38.66N

 

[cepheid]

First of all, both angles seem to be measured from the vertical. Therefore, I think that the vertical components of the tensions are scaled by cosine factors, and the horizontal components are scaled by sine factors. You should draw a right angle triangle representing the force vector resolved into its components (one for each tension) in order to be sure.