I made a pretty cool formula does it exist anywhere else? So I’ve made a formula that takes 2 angles and a base and depending on which angle you put in first gives you one of the 2 other sides and by using certain parts of the formula and it’s inverses you can solve pretty much any triangle problem (I think) I’d like to see if this formula already exists with that being said here’s the formula

edit#1: Hey I’m seeing a lot of downvotes on this post I’m a little confused on why please explain in the comments! That would be supper great of you thanks and have a good day!
edit#2: here is the work I did to get this formula it’s a little messy but overall it gives a good idea on how I made it https://docs.google.com/file/d/10w19VOlA_dAIrL6ZFsB6pq7sd1JdffhB/edit?usp=docslist_api&filetype=msword
 A: Let's simplify your formula:
$$h=\left(\sqrt{\left(\tan\left(\theta_1\frac{\pi}{180}+3\pi\right)\right)^2+1}\right)\left(\left(\frac{\cot\left(\theta_1\frac{\pi}{180}\right)}{\left(\cot\left(\theta_1\frac{\pi}{180}\right)+\cot\left(\theta_2\frac{\pi}{180}\right)\right)}\right)b\right)$$
First, let's agree that either all angles are in degrees and the trigonometric functions are calculated with the argument in degrees, or all angles are in radians and the trigonometric functions are calculated with the argument in radians: avoid mixing. The resulting formula doesn't need the factor $\frac{\pi}{180}$ anymore:
$$h=\left(\sqrt{\left(\tan(\theta_1+3\pi)\right)^2+1}\right)\left(\left(\frac{\cot\theta_1}{\left(\cot\theta_1+\cot\theta_2\right)}\right)b\right)$$
Next thing: $\tan(x+3\pi)=\tan x$ so we don't need $3\pi$ (also dispose of some unnecessary pairs of parentheses):
$$h=\left(\sqrt{\tan^2\theta_1+1}\right)\left(\frac{\cot\theta_1}{\cot\theta_1+\cot\theta_2}\right)b$$
Now, $\tan^2 x+1=\frac{\sin^2x}{\cos^2x}+1=\frac{\sin^2x}{\cos^2x}+\frac{\cos^2x}{\cos^2x}=\frac{1}{\cos^2x}$, so $\sqrt{\tan^2x+1}=\frac{1}{|\cos x|}$. This lets us transform this further:
$$h=\frac{1}{|\cos\theta_1|}\left(\frac{\cot\theta_1}{\cot\theta_1+\cot\theta_2}\right)b$$
The next thing we should do is write all the $\cot$'s as $\frac{\cos}{\sin}$:
$$h=\frac{1}{|\cos\theta_1|}\left(\frac{\frac{\cos\theta_1}{\sin\theta_1}}{\frac{\cos\theta_1}{\sin\theta_1}+\frac{\cos\theta_2}{\sin\theta_2}}\right)b$$
which is the same as:
$$h=\frac{1}{|\cos\theta_1|}\left(\frac{\cos\theta_1\sin\theta_2}{\cos\theta_1\sin\theta_2+\cos\theta_2\sin\theta_1}\right)b$$
or:
$$h=\frac{\cos\theta_1}{|\cos\theta_1|}\left(\frac{\sin\theta_2}{\sin(\theta_1+\theta_2)}\right)b$$
or:
$$h=\text{sgn}(\cos\theta_1)\times b\frac{\sin\theta_2}{\sin(\theta_1+\theta_2)}$$
where $\text{sgn}\,x=\begin{cases}1&x>0\\0&x=0\\-1&x<0\end{cases}$, and $\text{sgn}(\cos\theta_1)$ is in fact what you get when you cancel $\cos\theta_1$ with $|\cos\theta_1|$ in the previous line.
The last step is: if the third angle in the triangle is $\theta_3$, then $\theta_3=\pi-(\theta_1+\theta_2)$ and $\sin\theta_3=\sin(\theta_1+\theta_2)$, so your formula gives us:
$$h=\text{sgn}(\cos\theta_1)\times b \frac{\sin\theta_2}{\sin\theta_3}$$
Now, knowing the Law of sines, we can see that this is true up to the sign: the Law of sines states that:
$$\frac{h}{\sin\theta_2}=\frac{b}{\sin\theta_3}$$
i.e.
$$h=b \frac{\sin\theta_2}{\sin\theta_3}$$
which means that your formula is almost correct. The only problem with it is (apart from not working at all for $\theta_1=\pi/2$) that, for $\pi/2<\theta_1<\pi$ (where $\cos\theta_1$ is negative) it gives the result with the opposite sign of what it should be. Otherwise, it is a correct, albeit long-winded, expression equivalent to the Law of sines.
