This is an exercise I came across while tutoring high school physics. I am posting this as an "answer my own question."
Kyle suspends a 12340 N moose from two trees as shown below. What is the tension in the rope on the right?
The answer is shown below.
Here is a clip from my notebook:
There are three forces in equilibrium acting on the moose: It's weight of $12340 N$, and the tensions on the ropes, $T_L$ and $T_R$. The horizontal and vertical components of the tensions are also shown.
Since the body is in equilibrium, the net force is zero. All notation refers to the magnitudes of the vectors. Angles are in degrees. Forces and Tensions are in Newtons
The left and right horizontal forces are equal in magnitude:
$$F_H = T_R \cos 64 = T_L \cos 41$$
The sum of the upward vertical forces equal the magnitude of the downward vertical force: $$F_UR + F_UL = 12340$$
Combining terms to solve for $T_R$: $$F_UR = T_R \sin 64 $$ $$F_UL = T_L \sin 41 $$
Therefore: $$T_R \sin 64 + T_L \sin 41 = 12340$$ $$T_R \sin 64 + (\frac {F_H}{ \cos 41})\sin 41 = 12340$$ $$T_R \sin 64 + (\frac {T_R \cos 64}{ \cos 41})\sin 41 = 12340$$ $$T_R (\sin 64 + (\frac {\cos 64}{ \cos 41})\sin 41) = 12340$$
Finally:
$$T_R = \frac{12340}{\sin 64 + (\frac {\cos 64}{ \cos 41})\sin 41}$$ $$T_R \approx 9641.65 $$