Differentiating The Law of Cosines Alright, I've got a stupid question (maybe). I took the equation given by the law of cosines and I differentiated it with respect to some other parameter.
Right, so we begin be considering a triangle with vertices $A,B,$ and $C$. In $\triangle ABC,$ we have $|AB|=a,|BC|=b,$ and $|CA|=c.$ The angle opposite to $a$ is $\alpha,$ the angle opposite to $b$ is $\beta,$ and the angle opposite to $c$ is $\gamma.$
I begin by considering the cosine rule. We have that $a^2+b^2=c^2+2ab\cos{(\gamma)}.$ Now, this is the weird-ish part. I differentiated with respect to some other parameter, say $t$. So, we get $2[a\frac{da}{dt}+b\frac{db}{dt}]=2[c\frac{dc}{dt}-ab\sin{(\gamma)}\frac{d\gamma}{dt}+a\cos{(\gamma)}\frac{db}{dt}+b\cos{(\gamma)}\frac{da}{dt}].$ Cumbersome, I know. But, consider what happens when the sides $a,b,$ and $c$ don't change as $t$ changes. Most of the derivatives vanish, and we get $-ab\sin{(\gamma)}\frac{d\gamma}{dt}=0.$ Now, for a triangle, $a,b,$ and $\sin{(\gamma)}$ cannot be $0.$ This means that the angle $\gamma$ remains constant as $t$ varies. Similar arguments can be used for $\alpha,$ and $\beta.$ I know that I simply could've used $SSS$ congruency or something like that to prove that if all $3$ sides of a triangle remain constant, then so do the angles, but is this line of reasoning correct? Further, if we do something similar with the law of sines, we don't get a neat argument that shows that the angles remain constant as the sides do. Is there any way to rectify that? In particular, we get $\tan{(\beta)}\frac{d\alpha}{dt}=\tan{(\alpha)}\frac{d\beta}{dt},$ when the sine rule is applied to $a,b,\alpha,$ and $\beta,$ and differentiated. Nothing about this tells me that either of the derivatives are $0$.
 A: What you write is correct formally, but yours is not a valid argument to prove the SSS congruency criterion (if two triangles have the same sides, then they have the same angles, too).
In theory, it is possible that there are different configurations of the angle, but they cannot be turned one into the other via continuous deformations; you did not exclude this.
For instance, there are two different configurations for a triangle with fixed $a$, $b$, and $\sin \gamma$; one with an acute angle and one with an obtuse angle in $\gamma$. They have different lengths of the third side $c$. Both are rigid, so you cannot turn continuously one into the other while leaving $a$, $b$, $\sin\gamma$ unchanged. You should exclude that something similar happens in your case.
A: Rearrange the Law of Sines to avoid using the quotient rule:
$$b\sin\alpha = a\sin\beta,c\sin\alpha = a\sin\gamma, c\sin\beta = b\sin\gamma.$$
Differentiating, assuming the side lengths are constant, gives:
\begin{align}
b\cos(\alpha)\frac{d\alpha}{dt} &= a\cos(\beta)\frac{d\beta}{dt},\\
c\cos(\alpha)\frac{d\alpha}{dt} &= a\cos(\gamma)\frac{d\gamma}{dt},\\
c\cos(\beta)\frac{d\beta}{dt} &= b\cos(\gamma)\frac{d\gamma}{dt}.\\
\end{align}
Applying the Law of Sines, we can rewrite this as (*):
\begin{align}
\cot(\alpha)\frac{d\alpha}{dt} &= \cot(\beta)\frac{d\beta}{dt},\\
\cot(\alpha)\frac{d\alpha}{dt} &= \cot(\gamma)\frac{d\gamma}{dt},\\
\cot(\beta)\frac{d\beta}{dt} &= \cot(\gamma)\frac{d\gamma}{dt}.\\
\end{align}
From here, it's necessary to introduce additional constraints, e.g. that $\alpha+\beta+\gamma=\pi$, since otherwise there can be non-trivial solutions to this system. Differentiating, we get $\frac{d\alpha}{dt} + \frac{d\beta}{dt} + \frac{d\gamma}{dt} = 0.$ (**)
If all the angles are acute, we're done, since (*) implies all the derivatives have the same sign, which contradicts the fact that the sum of the derivatives is zero.
If one of the angles is a right angle, it follows from (*) that the other two derivatives are zero, which in turn implies all three derivatives are zero by (**).
Assume WLOG that $\gamma$ is an obtuse angle initially: $\alpha(0)\leq\beta(0)<\gamma(0)$ and $\gamma(0)> \frac{\pi}{2}$, so that $0<\alpha(0)+\beta(0)<\frac{\pi}{2}$.
Rewriting $\gamma$ and $\frac{d\gamma}{dt}$ using the above, we have:
\begin{align*}
\cot(\alpha)\frac{d\alpha}{dt} &= \cot(\alpha+\beta)\left(\frac{d\alpha}{dt}+\frac{d\beta}{dt}\right),\\
\cot(\beta)\frac{d\beta}{dt} &= \cot(\alpha+\beta)\left(\frac{d\alpha}{dt}+\frac{d\beta}{dt}\right),
\end{align*}
or
\begin{align*}
\cot(\alpha)\frac{d\alpha}{dt} &= \cot(\alpha+\beta)\left(\frac{d\alpha}{dt}+\frac{d\beta}{dt}\right) =  \cot(\alpha+\beta)(\tan(\alpha)\cot(\beta)+1)\frac{d\beta}{dt},\\
\cot(\beta)\frac{d\beta}{dt} &= \cot(\alpha+\beta)\left(\frac{d\alpha}{dt}+\frac{d\beta}{dt}\right)= \cot(\alpha+\beta)(\tan(\beta)\cot(\alpha)+1)\frac{d\alpha}{dt}.\\
\end{align*}
If both $\frac{d\alpha}{dt}$ and $\frac{d\beta}{dt}\neq 0 $, then matching with (*) we see that we must have:
$$
\cot(\beta) = \cot(\alpha+\beta)(\tan(\alpha)\cot(\beta)+1),
$$
and
$$
\cot(\alpha) = \cot(\alpha+\beta)(\tan(\beta)\cot(\alpha)+1),
$$
i.e.
$$
\tan(\alpha+\beta) = \tan(\alpha)+\tan(\beta).
$$
This has no solutions for $0 < \alpha\leq \beta < \frac{\pi}{2}$, $0<\alpha+\beta<\frac{\pi}{2}$, a contradiction.
