The following problem looks like it should be easy, but I don't know how to prove it rigorously. All I know is the sine rule should be applied somewhere.
Let $ABC$ be triangle with angles $\alpha$, $\beta$ and $\gamma$ and corresponding sides $a,b,c.$ Suppose that $\beta'$ and $\gamma'$ are angles such that $\alpha + \beta' + \gamma' = \pi$ and $\dfrac{b}{\sin{\beta'}} = \dfrac{c}{\sin{\gamma'}}$. Prove that $\beta' = \beta$ and $\gamma' = \gamma$.
I tried doing a proof by contradiction. Assume without loss of generality that $\beta' > \beta$. Then $\gamma' < \gamma$.
Now because of the sine rule: $\frac{b}{\sin{\beta}} = \frac{c}{\sin{\gamma}}$
And the statement of the problem tells us that $\frac{b}{\sin{\beta'}} = \frac{c}{\sin{\gamma'}}$.
Thus, $\frac{\sin{\beta}}{\sin{\gamma}} = \frac{\sin{\beta'}}{\sin{\gamma'}}$. Now since $\beta' > \beta$, then $\sin{\beta'} > \sin{\beta}$, and since $\gamma' < \gamma$, then $\sin{\gamma'} < \sin{\gamma}$. But if this is true, then $\frac{\sin{\beta}}{\sin{\gamma}} < \frac{\sin{\beta'}}{\sin{\gamma'}}$, which is a contradiction with what we said before.
Therefore, $\beta = \beta'$ and thus $\gamma = \gamma'$.
But the problem with this proof is that $\sin{\beta'}$ is not necessarily greater than $\sin{\beta}$, if $\beta'$ is greater than $\pi/2$ for example.
From $\frac{\sin{\beta}}{\sin{\gamma}} = \frac{\sin{\beta'}}{\sin{\gamma'}}$, using $\pi = \alpha + \beta +\gamma = \alpha + \beta' + \gamma'$,
$$\begin{align*} \frac{\sin{\beta}}{\sin{\gamma}} &= \frac{\sin{\beta'}}{\sin{\gamma'}}\\ \frac{\sin(\pi-\alpha - \gamma)}{\sin{\gamma}} &= \frac{\sin(\pi-\alpha - \gamma')}{\sin{\gamma'}}\\ \frac{\sin(\alpha + \gamma)}{\sin{\gamma}} &= \frac{\sin(\alpha + \gamma')}{\sin{\gamma'}}\\ \frac{\sin\alpha\cos\gamma+\cos\alpha\sin\gamma}{\sin{\gamma}} &= \frac{\sin\alpha\cos\gamma'+\cos\alpha\sin\gamma'}{\sin{\gamma'}}\\ \sin\alpha\cot\gamma+\cos\alpha &= \sin\alpha\cot\gamma'+\cos\alpha\\ \cot\gamma &= \cot\gamma'\\ \gamma &= \gamma' \end{align*}$$
Assuming all $\alpha, \beta,\gamma,\beta',\gamma'$ are strictly between $0$ and $\pi$, as mentioned in the comments.