## Task B-3.2.

In a triangle with sides of length $a, b, c$ and area $P$, the following equality holds:

$$
\sqrt{3}\left(b^{2}+a^{2}-c^{2}\right)=2 a b-4 P
$$

Calculate the measure of the angle opposite the side of length $c$.
## Solution.

Let the angle opposite the side of length $c$ be $\gamma$.

If we divide the equality $\sqrt{3}\left(b^{2}+a^{2}-c^{2}\right)=2 a b-4 P$ by $2 a b$, we get

$$
\frac{\sqrt{3}\left(b^{2}+a^{2}-c^{2}\right)}{2 a b}=1-\frac{2 P}{a b}
$$

or $\sqrt{3} \cos \gamma=1-\frac{2 P}{a b}$.

Since $P=\frac{1}{2} a b \sin \gamma$, from the previous equality we get the trigonometric equation $\sqrt{3} \cos \gamma=1-\sin \gamma$ which we solve by reducing to addition formulas:

$$
\begin{gathered}
\sqrt{3} \cos \gamma+\sin \gamma=1 \\
\frac{\sqrt{3}}{2} \cos \gamma+\frac{1}{2} \sin \gamma=\frac{1}{2} \\
\cos \gamma \cos 30^{\circ}+\sin \gamma \sin 30^{\circ}=\frac{1}{2} \\
\cos \left(\gamma-30^{\circ}\right)=\frac{1}{2}
\end{gathered}
$$

From $\cos \left(\gamma-30^{\circ}\right)=\frac{1}{2}$ it follows that $\gamma-30^{\circ}=60^{\circ}$, or $\gamma=90^{\circ}$.

Note: The equation $\sqrt{3} \cos \gamma+\sin \gamma=1$ can also be solved using a universal substitution. By introducing the substitution $\operatorname{tg} \frac{\gamma}{2}=t$, the equation $\sqrt{3} \cos \gamma+\sin \gamma=1$ transforms into the equation

$$
\sqrt{3} \frac{1-t^{2}}{1+t^{2}}+\frac{2 t}{1+t^{2}}=1
$$

or $(1+\sqrt{3}) t^{2}-2 t+1-\sqrt{3}=0$, and the solutions are $t_{1}=\sqrt{3}-2$ and $t_{2}=1$.

From $\operatorname{tg} \frac{\gamma}{2}=\sqrt{3}-2180^{\circ}$ which is impossible.

From $\operatorname{tg} \frac{\gamma}{2}=1$ it follows that $\gamma=90^{\circ}$.