#### Answer

The tension in the third rope is 94.3 N and it points at an angle of $58.0^{\circ}$ below the horizontal.

#### Work Step by Step

The vertical component of $T_3$ must be equal in magnitude to $T_2$. The horizontal component of $T_3$ must be equal in magnitude to $T_1$. We can find the magnitude of $T_3$ as;
$T_3 = \sqrt{(T_{3y})^2+(T_{3x})^2}$
$T_3 = \sqrt{(80~N)^2+(50~N)^2}$
$T_3 = 94.3~N$
We then find the angle $\theta$ below the horizontal;
$tan(\theta) = \frac{T_{3y}}{T_{3x}}$
$\theta = arctan(\frac{80~N}{50~N})$
$\theta = 58.0^{\circ}$
The tension in the third rope is 94.3 N and it points at an angle of $58.0^{\circ}$ below the horizontal.