[Geometry]
Problem: Triangle $A B C$ has $\mathrm{m} \angle A>\mathrm{m} \angle B>\mathrm{m} \angle C$. The angle between the altitude and the angle bisector at vertex $A$ is $6^{\circ}$. The angle between the altitude and the angle bisector at vertex $B$ is $18^{\circ}$. Compute the degree measure of angle $C$.

Solution: Let the feet of the altitudes from $A$ and $B$ be $E$ and $D$, respectively, and let $F$ and $G$ be the intersection points of the angle bisectors with $\overline{A C}$ and $\overline{B C}$, respectively, as shown below.

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Then $\mathrm{m} \angle G A E=6^{\circ}$ and $\mathrm{m} \angle D B F=18^{\circ}$. Suppose $\mathrm{m} \angle F B C=x^{\circ}$ and $\mathrm{m} \angle C A G=y^{\circ}$. So $\mathrm{m} \angle C A E=(y+6)^{\circ}$ and $\mathrm{m} \angle C B D=(x+18)^{\circ}$. Considering right triangle $B D C$, $\mathrm{m} \angle C=90^{\circ}-(x+18)^{\circ}=(72-x)^{\circ}$, while considering right triangle $A E C, \mathrm{~m} \angle C=$ $90^{\circ}-(y+6)^{\circ}=(84-y)^{\circ}$. Thus $84-y=72-x$ and $y-x=12$. Considering $\triangle A B E$, $\mathrm{m} \angle E A B=(y-6)^{\circ}$ and $\mathrm{m} \angle E B A=2 x^{\circ}$, so $(y-6)+2 x=90$, or $2 x+y=96$. Solving the system yields $x=28, y=40$. Therefore $\mathrm{m} \angle A=80^{\circ}$ and $\mathrm{m} \angle B=56^{\circ}$, so $\mathrm{m} \angle C=44^{\circ}$.
From right triangle $A B E, 90^{\circ}=\left(\frac{1}{2} A-6^{\circ}\right)+B$, and from right triangle $A B D, 90^{\circ}=\left(\frac{1}{2} B-18^{\circ}\right)+A$. Adding the two equations gives $180^{\circ}=\frac{3}{2}(A+B)-24^{\circ}$, so $A+B=\frac{2}{3} \cdot 204^{\circ}=136^{\circ}$ and $C=180^{\circ}-(A+B)=44^{\circ}$.

Final answer: $44^{\circ}$