[Geometry]
Problem: Suppose that noncongruent triangles $A B C$ and $X Y Z$ are given such that $A B=X Y=10, B C=$ $Y Z=9$, and $\mathrm{m} \angle C A B=\mathrm{m} \angle Z X Y=30^{\circ}$. Compute $[A B C]+[X Y Z]$.

Solution: Because triangles $A B C$ and $X Y Z$ are noncongruent yet have two adjacent sides and an angle in common, the two triangles are the two possibilities in the ambiguous case of the Law of Sines. Without loss of generality, let triangle $A B C$ have obtuse angle $C$ and triangle $X Y Z$ have acute angle $Z$ so that $\mathrm{m} \angle C+\mathrm{m} \angle Z=$ $180^{\circ}$. Place triangle $A B C$ so that $B$ and $Y$ coincide, and $C$ and $Z$ coincide. Because $\mathrm{m} \angle C$ and $\mathrm{m} \angle Z$ add up to $180^{\circ}$, it follows that points $X, Z$, and $A$ all lie on the same line. The two triangles together then form $\triangle A B X$, where $\mathrm{m} \angle B A X=\mathrm{m} \angle B X A=30^{\circ}$ and $B X=A B=10$. Therefore the sum of the areas of the two triangles is equal to the area of triangle $A B X$, which is $\frac{1}{2} \cdot 10 \cdot 10 \cdot \sin \left(120^{\circ}\right)=\frac{5 \cdot 10 \cdot \sqrt{3}}{2}=\mathbf{2 5} \sqrt{\mathbf{3}}$.

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Figure not drawn to scale.
As explained above, let $\triangle A B C$ have obtuse angle $C$ and $\triangle X Y Z$ have acute angle $Z$. By the Law of Sines, $\sin (\angle C)=\sin (\angle Z)=\frac{5}{9}$. This implies $\mathrm{m} \angle X Y Z=\frac{5 \pi}{6}-\arcsin \left(\frac{5}{9}\right)$ and $\mathrm{m} \angle A B C=$ $\arcsin \left(\frac{5}{9}\right)-\frac{\pi}{6}$. The areas of the triangles are $[X Y Z]=\frac{1}{2} \cdot 10 \cdot 9 \cdot \sin \left(\frac{5 \pi}{6}-\arcsin \left(\frac{5}{9}\right)\right)$ and $[A B C]=\frac{1}{2} \cdot 10 \cdot 9$. $\sin \left(\arcsin \left(\frac{5}{9}\right)-\frac{\pi}{6}\right)$. By the angle subtraction rule, it follows that

$$
\begin{aligned}
\sin \left(\frac{5 \pi}{6}-\arcsin \left(\frac{5}{9}\right)\right) & =\sin \left(\frac{5 \pi}{6}\right) \cos \left(\arcsin \left(\frac{5}{9}\right)\right)-\cos \left(\frac{5 \pi}{6}\right) \sin \left(\arcsin \left(\frac{5}{9}\right)\right) \text { and } \\
\sin \left(\arcsin \left(\frac{5}{9}\right)-\frac{\pi}{6}\right) & =\sin \left(\arcsin \left(\frac{5}{9}\right)\right) \cos \left(\frac{\pi}{6}\right)-\cos \left(\arcsin \left(\frac{5}{9}\right)\right) \sin \left(\frac{\pi}{6}\right) .
\end{aligned}
$$

The sum of the two sines is $\sin \left(\arcsin \left(\frac{5}{9}\right)\right)\left(\cos \left(\frac{\pi}{6}\right)-\cos \left(\frac{5 \pi}{6}\right)\right)=\frac{5}{9} \cdot \sqrt{3}$ because $\sin \left(\frac{\pi}{6}\right)=\sin \left(\frac{5 \pi}{6}\right)$. Finally, the sum of the areas of the two triangles is $\frac{1}{2} \cdot 10 \cdot 9 \cdot \frac{5}{9} \cdot \sqrt{3}=25 \sqrt{3}$.

Final answer: $25 \sqrt{3}$