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, 9 months ago

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– Akshat Sharda · 1 year, 9 months ago

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

– Brian Wang · 1 year, 9 months ago

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

– Akshat Sharda · 1 year, 9 months 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, 9 months ago

ah, okLog in to reply