8 The three sides of a triangle are $t^{2}+t+1$, $t^{2}-1$, and $2 t+1$, then the largest interior angle of the triangle is $\qquad$ .
8 Given that the side lengths are positive, we know $t>1$. It is easy to see that $t^{2}+t+1$ is the largest number, and
$$
t^{2}+t+1<t^{2}+2 t=\left(t^{2}-1\right)+(2 t+1),
$$

thus a triangle can be formed with sides $t^{2}+t+1$, $t^{2}-1$, and $2 t+1$. If $A$ is the largest angle, then
$$
\begin{aligned}
\cos A & =\frac{\left(t^{2}-1\right)^{2}+(2 t+1)^{2}-\left(t^{2}+t+1\right)^{2}}{2\left(t^{2}-1\right)(2 t+1)} \\
& =-\frac{1}{2},
\end{aligned}
$$

so $A=120^{\circ}$.