8. (10 points) In the inscribed quadrilateral $A B C D$, the degree measures of the angles are in the ratio $\angle A: \angle B: \angle C=2: 3: 4$. Find the length of $A C$, if $C D=15, B C=18 \sqrt{3}-7.5$.
Answer: 39

## Solution:

![](https://cdn.mathpix.com/cropped/2024_05_06_61450ad4ec176109d5e0g-19.jpg?height=848&width=531&top_left_y=341&top_left_x=754)

The quadrilateral is inscribed, hence $\angle A+\angle C=180^{\circ}$. From the given ratio, $\angle A=2 x, \angle C=4 x$. Therefore, $x=30^{\circ}$ and $\angle A=60^{\circ}, \angle B=3 x=90^{\circ}$. Extend $A B$ and $C D$ until they intersect at point $K$. Then $\angle K=30^{\circ}, K C=2 B C=36 \sqrt{3}-15, K D=K C+C D=36 \sqrt{3}$ and $A D=36$. Using the Pythagorean theorem, we get $A C=39$.

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