I have a conundrum on my hands. The simple question is this:
Given isosceles triangle $ABC$, where $m\angle B$ is equal to $m\angle C$, and where $m\angle A$ is equal to $x+12$, what is the value of $x$?
Within the context of the original prompt, we can assume that $m\angle B$ is equal to $\frac{180-(x+12)}{2}$. That means that $m∠B+m∠C=180-(x+12)$. Let us refer to $m∠B+m∠C$ as $k$ for now. Simplified, $k=-x+168$.
To find the value of $x$, then, we should infer that $k+m∠A=180$. Let's do that. Now we have this: $-x+168+x+12=180$ (the $x+12$ came from the value of $m\angle A$), before simplification, since if we do so, we run into a problem. $-x$ and $x$ cancel out, leaving $168+12=180$, a true statement, yet not very helpful.
Let's take a step back and simplify only the integers. Now we have $-x+x+180=180$. My first instinct was that the answer was $0$, but upon closer inspection, I realized that the answer can be any real number, yet the original problem asked for a definite answer. After hounding down the answer sheet, I can see that the answer was $13$. While this works ($-13+13+180=180$, after all), it still dumbfounds me as to why the answer was $13$, especially since this wasn't multiple choice. Am I missing something here?