Answer
The tension in the cable B is $~~3650~N$
Work Step by Step
The horizontal component of $T_C$ is equal in magnitude to the horizontal component of $T_B$:
$T_C~cos~\theta_2 = T_B~cos~\theta_1$
$T_C = \frac{T_B~cos~\theta_1}{cos~\theta_2}$
The sum of the vertical components of $T_B$ and $T_C$ is equal in magnitude to the weight of the sphere:
$T_C~sin~\theta_2+T_B~sin~\theta_1 = T_A$
$\frac{T_B~cos~\theta_1}{cos~\theta_2}~sin~\theta_2+T_B~sin~\theta_1 = T_A$
$T_B~cos~\theta_1~tan~\theta_2+T_B~sin~\theta_1 = T_A$
$T_B = \frac{T_A}{cos~\theta_1~tan~\theta_2+sin~\theta_1}$
$T_B = \frac{(817~kg)(9.8~m/s^2)}{cos~51.0^{\circ}~tan~66.0^{\circ}+sin~51.0^{\circ}}$
$T_B = 3650~N$
The tension in the cable B is $~~3650~N$
