Answer
The tension in the rope to the left is $926N$ and the one to the right is $860N$
Work Step by Step
Let's call the tension in the rope to the left $T_l$ and the tension in the rope to the right $T_r$. The climber is also under the influence of her weight $W=535N$.
Since the climber rests and therefore is motionless at the point, we have $$\vec{W}+\vec{T}_l+\vec{T}_r=0 (1) $$
We will solve this exercise by using unit-vector notation. Taking rightward to be $+x$ direction and upward to be $+y$ direction, we rewrite the forces:
- The climber's weight (points downward): $\vec{W}=-535Nj$
- Tension $T_l$ (points leftward and upward): $\vec{T}_l=-(T_l\sin65N)i+(T_l\cos65N)j=(-0.91T_lN)i+(0.42T_l)j$
- Tension $T_r$ (points rightward and upward): $\vec{T}_r=(T_r\sin80N)i+(T_r\cos80N)j=(0.98T_rN)i+(0.17T_rN)j$
From equation (1), we have
$-0.91T_l+0.98T_r=0$
$0.42T_l+0.17T_r-535=0$
Solve for $T_l$ and $T_r$, we get $T_l=926N$ and $T_r=860N$
