Cosine Theorem

A circle can be circumscribed around quadrilateral $ABCD$. In addition, $AB=3, BC=4, CD=5$ and $AD=2$.

Find $AC$.
Apply the Law of Cosines and the property of a cyclic quadrilateral.

## Solution

Let $\angle A B C=\alpha$. Then

$$
A C^{2}=A B^{2}+B C^{2}-2 A B \cdot B C \cos \alpha=A D^{2}+C D^{2}-2 A D \cdot C D \cos \left(180^{\circ}-\alpha\right)
$$

or

$$
9+16-2 \cdot 3 \cdot 4 \cos \alpha=4+25+2 \cdot 2 \cdot 5 \cos \alpha
$$

From this equation, we find that $\cos \alpha=-\frac{1}{11}$. Therefore,

$$
A C^{2}=9+16+2 \cdot 3 \cdot 4 \cdot \frac{1}{11}=\frac{299}{11}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_36c4956010bd91c4b4eeg-05.jpg?height=403&width=412&top_left_y=2331&top_left_x=823)

## Answer

Problem