In triangle $A B C$, side $A B=6, \angle B A C=30^{\circ}$, the radius of the circumscribed circle is 5. Find side $A C$.

#
Apply the generalized sine theorem and the cosine theorem.

## Solution

Let $R$ be the radius of the circumscribed circle of triangle $ABC$. Then

$$
BC = 2R \sin \angle BAC = 2R \sin 30^\circ = R = 5
$$

Let $AC = x$. By the cosine theorem,

$$
BC^2 = AB^2 + AC^2 - 2AB \cdot AC \cos 30^\circ, \text{ or } 25 = 36 + x^2 - 6x \sqrt{3}
$$

From this equation, we find that $x = 3\sqrt{3} \pm 4$. For each found $x$, the sum of the two smaller sides of the triangle is greater than the third side. Therefore, $AC = 3\sqrt{3} + 4$ or $AC = 3\sqrt{3} - 4$.

## Answer

$AC = 3\sqrt{3} \pm 4$.