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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 \(\frac{(k \sin \gamma + n \sin \alpha + m \sin \beta)^{2}}{2 \sin \alpha \sin \beta \sin \gamma}\).

Note by Akshat Sharda
11 months, 2 weeks ago

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The question is confusing. I believe you missed a number "Prove the area of a triangle with angles x,y,z knowing that" Brian Wang · 11 months, 2 weeks ago

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@Brian Wang No !! its correct. Please tell me what are you confused at. Akshat Sharda · 11 months, 2 weeks ago

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@Akshat Sharda what is the area I'm supposed to prove? Brian Wang · 11 months, 2 weeks ago

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@Brian Wang You have to prove that area of \(\triangle ABC =\frac{(k \sin \gamma + n \sin \alpha + m \sin \beta)^{2}}{2 \sin \alpha \sin \beta \sin \gamma}\) Akshat Sharda · 11 months, 2 weeks ago

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@Akshat Sharda ah, ok Brian Wang · 11 months, 2 weeks ago

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