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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
2 years, 1 month 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 - 2 years, 1 month ago

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No !! its correct. Please tell me what are you confused at.

Akshat Sharda - 2 years, 1 month ago

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what is the area I'm supposed to prove?

Brian Wang - 2 years, 1 month 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 - 2 years, 1 month ago

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@Akshat Sharda ah, ok

Brian Wang - 2 years, 1 month ago

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