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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TopNewestThe 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, 1 month ago

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– Akshat Sharda · 1 year, 1 month ago

No !! its correct. Please tell me what are you confused at.Log in to reply

– Brian Wang · 1 year, 1 month ago

what is the area I'm supposed to prove?Log in to reply

– Akshat Sharda · 1 year, 1 month ago

You have to prove that area of \(\triangle ABC =\frac{(k \sin \gamma + n \sin \alpha + m \sin \beta)^{2}}{2 \sin \alpha \sin \beta \sin \gamma}\)Log in to reply

– Brian Wang · 1 year, 1 month ago

ah, okLog in to reply