[Geometry]
Problem: Let $T=95$. Triangle $A B C$ has $A B=A C$. Points $M$ and $N$ lie on $\overline{B C}$ such that $\overline{A M}$ and $\overline{A N}$ trisect $\angle B A C$, with $M$ closer to $C$. If $\mathrm{m} \angle A M C=T^{\circ}$, then $\mathrm{m} \angle A C B=U^{\circ}$. Compute $U$.

Solution: Because $\triangle A B C$ is isosceles with $A B=A C, \mathrm{~m} \angle A B C=U^{\circ}$ and $\mathrm{m} \angle B A C=(180-2 U)^{\circ}$. Therefore $\mathrm{m} \angle M A C=\left(\frac{180-2 U}{3}\right)^{\circ}=\left(60-\frac{2}{3} U\right)^{\circ}$. Then $\left(60-\frac{2}{3} U\right)+U+T=180$, so $\frac{1}{3} U=$ $120-T$ and $U=3(120-T)$. Substituting $T=95$ yields $U=\mathbf{7 5}$.

Final answer: 75