Answer
(a) $T_B = 46,300~N$
(b) $T_A = 29,800~N$
Work Step by Step
(a) The vertical component of $T_B$ is equal in magnitude to the weight of the wrecking ball.
$T_B~cos(40^{\circ}) = mg$
$T_B = \frac{(3620~kg)(9.80~m/s^2)}{cos(40^{\circ})}$
$T_B = 46,300~N$
(b) The tension $T_A$ is equal in magnitude to the horizontal component of $T_B$.
$T_A = T_B~sin(40^{\circ})$
$T_A = (46,300~N)~sin(40^{\circ})$
$T_A = 29,800~N$
