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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Top NewestThe question is confusing. I believe you missed a number "Prove the area of a triangle with angles x,y,z knowing that" – Brian Wang · 1 year, 9 months ago
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No !! its correct. Please tell me what are you confused at. – Akshat Sharda · 1 year, 9 months ago
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what is the area I'm supposed to prove? – Brian Wang · 1 year, 9 months ago
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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 · 1 year, 9 months ago
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ah, ok – Brian Wang · 1 year, 9 months ago
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