Answer
(a) $T = 2540~N$
(b) The minimum angle is $\theta = 1.01^{\circ}$.
Work Step by Step
(a) The tension $T$ is equal in both sides of the rope. The sum of the vertical components in each side of the rope will be equal in magnitude to the person's weight.
$2T~sin(\theta) = mg$
$T = \frac{mg}{2~sin(\theta)}=\frac{(90.0~kg)(9.80~m/s^2)}{2~sin(10.0^{\circ})}$
$T = 2540~N$
(b) Let's suppose the tension in the rope is $2.50\times 10^4~N$.
$2T~sin(\theta) = mg$
$sin(\theta) = \frac{mg}{2T}$
$\theta =arcsin(\frac{(90.0~kg)(9.80~m/s^2)}{(2)(2.50\times 10^4~N)})$
$\theta = 1.01^{\circ}$
The minimum angle is $\theta = 1.01^{\circ}$.