Why can't I use $\tan()$ to find the force of tension within a rope that forms a triangle? Apologies if this is too simple, but my teachers could not give me an answer. The following is a question from my grade 12 physics homework:

Now, the solution for $T_2$ is simple:
$T_{1_x} =T_{2_x}$
$96 cos(60) = T_2cos(30)$
$T_2 = 96 cos(60) / cos(30)$
$T_2 = 55 N $
This is all logical, but this is a right angle triangle so the following should also work:
$tan(60) = T_2 / 96$
$96 tan(60) = T_2$
$T_2 = 166 N $
It does not. 55N is the correct answer. So, why can I not use tan() to find the value to $T_2$?
 A: Your intuition is correct, but you are using the wrong angle.
$$\tan(30°) = \frac{T_2}{96\ \mbox{N}} \Rightarrow T_2 \simeq 55\ \mbox{N}$$
Or, alternatively, you are using the wrong definition of tangent:
$$\tan(60°) = \frac{96\ \mbox{N}}{T_2} \Rightarrow T_2 \simeq 55\ \mbox{N}$$

I was too quick in my previous answer. I arrogantly assumed that this was a simple trigonometry mistake and I tried changing the angle on the calculator, which gave me the right result on the first try, further fueling my belief.
However, as noted by the OP, the definition of tangent is correct: opposite/adjacent. And yet here the opposite works.
Let's restart from this step in your calculation:
$$T_2 = 96\ \mbox{N}\ \frac{\cos(60°)}{\cos(30°)}$$
Now, if we remember that $\cos(60°) = \sin(30°)$, we get that $T_2 = 96\ \mbox{N} \tan(30°)$. Alternatively, we could use $\cos(30°) = \sin(60°)$ to obtain $T_2 = \frac{96\ \mbox{N}}{\tan(60°)}$.
This justifies the two equations I wrote above, but why is there an apparent conflict with the geometry of the problem?
A first hint comes from the "length" of the sides. Notice how the short side is $96\ \mbox{N}$ "long", while the long side is only $55\ \mbox{N}$ "long"!
This tells you that the geometry you see in the picture is valid to calculate the length of the wires holding what I assume is a frame, but it does not apply to the tensions. If you actually draw the tensions to scale, you will notice that the triangle they form has a vertical hypotenuse, instead of a horizontal one. That's because the resulting force you get by summing the two tensions has to be purely vertical to balance the weight of the frame.
Sorry for the inaccurate answer before.
A: You are implicitly assuming that the vertical components of the tensions in the two parts of the string will be equal, in which case
$T_1 \sin 60^o = T_2 \sin 30^o
\\ \Rightarrow T_1 \sin 60^o = T_2 \cos 60^o
\\ \Rightarrow T_2 = T_1 \tan 60^o$
But this is incorrect because it is the horizontal components of the tensions that are equal, not the vertical components. So we have
$T_1 \cos 60^o = T_2 \cos 30^o
\\ \Rightarrow T_1 \cos 60^o = T_2 \sin 60^o
\\ \displaystyle \Rightarrow T_2 = \frac {T_1}{\tan 60^o}$
A: The theory is correct, but your angles are reversed.  Notice that the given angles are from the horizontal.  The hypotenuse of the force triangle will be the weight, a vertical force.  $T_1$ is 60 degrees from the horizontal but only 30 degrees from the vertical.  Similarly, $T_2$ is 60 degrees from the vertical.  Start with a free body diagram with vectors in fairly accurate directions, then find the angles relative to the vertical, since the vertical direction is the hypotenuse of your forces added together.
