Answer
The tension in rope 3 is 66.6 N and the angle of rope 3 is $36.1^{\circ}$
Work Step by Step
Let $T_3$ be the tension in rope 3. Let $T_1$ be the tension in rope 1. The vertical component of $T_1$ is equal to the cat's weight.
$T_1~sin(\theta_1) = m_{cat}~g$
$T_1 = \frac{m_{cat}~g}{sin(\theta_1)}$
$T_1 = \frac{(2.0~kg)(9.80~m/s^2)}{sin(20^{\circ})}$
$T_1 = 57.3~N$
The horizontal component of the tension in rope 1 is equal to the tension in rope 2 which is equal to the horizontal component of the tension in rope 3.
$T_{3x} = T_1~cos(\theta_1)$
$T_{3x} = (57.3~N)~cos(20^{\circ})$
$T_{3x} = 53.8~N$
The vertical component of $T_3$ is equal to the dog's weight.
$T_{3y} = m_{dog}~g$
$T_{3y} = (4.0~kg)(9.80~m/s^2)$
$T_{3y} = 39.2~N$
We can find the tension $T_3$.
$T_3 = \sqrt{(T_{3x})^2+(T_{3y})^2}$
$T_3 = \sqrt{(53.8~N)^2+(39.2~N)^2}$
$T_3 = 66.6~N$
We can find the angle $\theta$ of rope 3.
$tan(\theta) = \frac{T_{3y}}{T_{3x}}$
$\theta = arctan(\frac{39.2~N}{53.8~N})$
$\theta = 36.1^{\circ}$
The tension in rope 3 is 66.6 N and the angle of rope 3 is $36.1^{\circ}$.
