Please take a look at the picture below for the diagram reference:
I am trying to calculate the point where it is perfectly 3.3 cm vertically from the 44.52 cm line AND 5.5 cm horizontally from the 12 cm line
(I'm using rhinoceros to model this, but the solution doesn't need to be via rhinoceros, can be any way)
Would really appreciate any help to achieve this!
This can be solved with analytic geometry, which is using algebra, especially graphing in the Cartesian plane, to solve geometry problems. We'll also need a little trigonometry, since you give some angles.
Let's place that figure on the Cartesian plane. We'll place the origin $(0,0)$ at the top corner of your figure. The $44.52$ line segment goes through the origin and has the angle of inclination $31.50°$, so its slope is $\tan 31.50°$. Your desired point is $3.30$ below that, so it is on the line with the same slope and $y$-intercept $-3.30$, namely
$$y=(\tan 31.50°)x-3.30$$
The $12.00$ line segment goes through the origin and has the angle of inclination $-43.50°$, so its slope is $\tan -43.50°$. Your desired point is $5.50$ to the left of that, so it is on the line with the same slope and $x$-intercept $-5.50$, namely
$$y=(\tan -43.50°)(x+5.50)$$
Solve those two simultaneous equations. The value of $x$ will be the distance to the right your desired point is from the top point: a negative value of $x$ means your point is actually to the left. The value of $y$ will be the distance up your desired point is from the top point: a negative value of $y$ means your point is actually down.
I'll leave the equation solving to you. Let me know if you have trouble with that. Those tangents means the values of $x$ and $y$ will be irrational numbers. You can decide if the exact values are needed or if good approximations will suffice.