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 = 67.5° and β = 75.0°.)

Given:
α = 67.5 ° , with VERTICAL, and
β = 75 ° with VERTICAL
weight of climber, W = 586 N
Need to find tensions of ropes.

Let
T1=tension to the left of climber
T2=tension to the right of climber

Solution:
A. draw FBD (free body diagram on the climber

B. using sum of horizontal forces = 0
-T1sin(alpha) +T2 sin(beta) = 0,
substitute known values
T1sin(67.5) = T2sin(75)
0.923880T1=0.99619T2
T1=1.07827*T2.......................................(1)

C. using sum of vertical forces = 0
T1cos(alpha)+T2cos(beta)-W=0
substitute known values,
T1cos(67.5)+T2cos(75)-586=0
0.38268T1+0.087156T2-586=0..................(2)

D. Solve system of equations (1) & (2)
substitute T1 from (1) into (2)

0.38268(1.07827*T2)+0.087156T2-586=0 =>
0.49979T2-586=0
T2=1172.485
Substitute into (1)
T1=1.078273T2=1264.260

E. Check solution

Equation (1):
T1sin(alpha)=1168.024
T2sin(beta) = 1168.024 ok.

Equation (2)
T1cos(alpha)+T2cos(alpha)
=483.811+102.189
=586.000 ok