## [ Law of Sines ]

 [Law of Cosines] Two sides of a triangle are equal to $a$ and $b$. Find the third side of the triangle if the angle opposite the third side is twice the angle opposite the side equal to $b$.

#
## First method.

Let $\alpha$ be the angle opposite the side equal to $b$. By the Law of Sines,

$$
\frac{a}{\sin \left(18 \sigma^{\gamma}-3 \times 2\right)}=\frac{a}{\sin 3(2)}=\frac{b}{\sin (2)}.
$$

Since

$$
\sin 3 \alpha=\sin \alpha(2 \cos 2 \alpha+1),
$$

then

$$
2 \cos 2 \alpha=\frac{a-b}{b}.
$$

Then, by the Law of Cosines,

$$
c^2=a^2+b^2-2ab \cos 2 \alpha=a^2+b^2-a(a-b)=b^2+ab=b(a+b).
$$

## Second method.

Let $\alpha$ be the angle opposite the side equal to $b$. If $l$ is the angle bisector of the angle opposite the side $c$ we are looking for, then

$$
l=\frac{2ab \cos \frac{\alpha}{2}}{a+b}.
$$

On the other hand, this angle bisector cuts off an isosceles triangle from the given triangle with base $a$ and angle at the base $\alpha$. Therefore,

$$
l=\frac{a}{2 \cos \frac{\alpha}{2}}.
$$

Thus,

$$
2 \cos 2 \alpha=\frac{a+b}{2b}, \cos 2 \alpha=2 \cos 2 \alpha-1=\frac{a-b}{2b}.
$$

Then, by the Law of Cosines,

$$
c^2=a^2+b^2-2ab \cos 2 \alpha=a^2+b^2-a(a-b)=b^2+ab=b(a+b).
$$

## Third method.

Draw the angle bisector $CD$ of triangle $ABC$, where $BC=a$, $AC=b$. Then $\angle ACD = \angle BCD = \frac{1}{2} \angle ACB = \angle ABC$, so triangle $CDB$ is isosceles, $CD=DB$, and triangles $ACD$ and $ABC$ are similar (by two angles). Therefore, the ratio of the sum of sides $AD$ and $DC$, enclosing the angle $ADC$ of triangle $ACD$, to the sum of sides $AC$ and $BC$, enclosing the angle $ACB$ of triangle $ABC$, is the similarity coefficient, i.e., the ratio of sides $AC$ and $AB$. Since $CD=DB$ and $AB=AD+DB=AD+DC$, then

$$
\frac{AD+DC}{AC+BC}=\frac{AC}{AB}, \frac{AB}{a+b}=\frac{b}{AB}.
$$

From this, we find that

$$
AB^2=b(a+b), AB=\sqrt{b(a+b)}.
$$

## Answer  $\sqrt{b(a+b)}$