An easy proof problem !!

Prove that the area of triangle with angles \(\alpha,\beta,\gamma\) knowing that the distances from an arbitrary point \(M\) taken inside the triangle to its sides are equal to \(m,n,k\) is equal to \(\frac{(k \sin \gamma + n \sin \alpha + m \sin \beta)^{2}}{2 \sin \alpha \sin \beta \sin \gamma}\).

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An easy proof problem !!

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