<s>[INST] What is wrong with my solution to find the size of an angle of a triangle? The sides of a triangle $a$, $b$, and $c$ has the following lengths.
$$a=x^2+x+1$$
$$b=2x+1$$
$$c=x^2-1$$
Since $a,b,c>0$, $x>1$.
$$a-b=x(x-1)>0\ (\because x>1)$$
$$a-c=x+2>0$$
Therefore, $a$ is the longest side, and its opposite angle $A$ is the largest angle. Let $B$ and $C$ be the opposite angle of $b$ and $c$, respectively.
$$a:b:c=x^2+x+1:2x+1:x^2-1=A:B:C$$
$$A=(x^2+x+1)k$$
$$B=(2x+1)k$$
$$C=(x^2-1)k$$
$$A+B+C=(2x^2+3x+1)k=\pi$$
$$k=\frac{\pi}{2x^2+3x+1}$$
$$A=\frac{(x^2+x+1)\pi}{2x^2+3x+1}$$
The correct value of A is $\frac{2}{3}\pi$, but the given solution doesn't produce the right answer. I got the correct value by an alternative method, but I can't figure out what's wrong with the solution above.
 [/INST] Well, the angles A,B,C are not "proportional" as you wrote.
By the cosine law, you have
$$
a^2=b^2+c^2-2bc \cos(A).
$$
Hence
$$
\cos(A)=\frac{b^2+c^2-a^2}{2bc}=\frac{(2x+1)^2+(x^2-1)^2-(x^2+x+1)^2}{2(2x+1)(x^2-1)}=-\frac{1}{2}.
$$
Therefore $A=\frac{2}{3}\pi$.
</s>