[Law of Cosines] [Law of Sines]

In a triangle, the sides are known as \(a=2\), \(b=3\), and the area \(S=\frac{3 \sqrt{15}}{4}\). The median drawn to its third side is less than half of it.

Find the radius of the circumscribed circle of this triangle.
Prove that the angle opposite the third side is obtuse and use the cosine and sine theorems.

## Solution

Let $\alpha$ be the angle of the triangle opposite the third side, equal to $c$. Then $\sin \alpha=2 S / a b=\frac{\sqrt{15}}{4}$. Since the median drawn to the side equal to $c$ is less than $c / 2$, the vertex of angle $\alpha$ of the given triangle lies inside the circle constructed on side $c$ as the diameter, hence, angle $\alpha$ is obtuse, and

$\cos \alpha=-\sqrt{1-\sin ^{2} \alpha}=-1 / 4$

By the cosine theorem $c^{2}=a^{2}+b^{2}-2 a b \cos \alpha=4+9+3=16$. If $R$ is the radius of the circumscribed circle, then $R=$ $c / 2 \sin \alpha=\frac{8}{\sqrt{15}}$

## Answer

$\frac{8}{\sqrt{15}}$.