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 $C D=5, B C=6 \sqrt{3}-2.5$.
Answer: 13

## Solution:

![](https://cdn.mathpix.com/cropped/2024_05_06_61450ad4ec176109d5e0g-27.jpg?height=848&width=531&top_left_y=341&top_left_x=754)

The quadrilateral is inscribed, so $\angle A + \angle C = 180^{\circ}$. From the given ratio, $\angle A = 2x, \angle C = 4x$. Therefore, $x = 30^{\circ}$ and $\angle A = 60^{\circ}, \angle B = 3x = 90^{\circ}$. Extend $AB$ and $CD$ to intersect at point $K$. Then $\angle K = 30^{\circ}, KC = 2BC = 12\sqrt{3} - 5, KD = KC + CD = 12\sqrt{3}$, and $AD = 12$. Using the Pythagorean theorem, we get $AC = 13$.

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