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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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## Comments

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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"

- 2 years, 1 month ago

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

- 2 years, 1 month ago

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what is the area I'm supposed to prove?

- 2 years, 1 month ago

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

- 2 years, 1 month ago

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ah, ok

- 2 years, 1 month ago

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