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 $\frac{(k \sin \gamma + n \sin \alpha + m \sin \beta)^{2}}{2 \sin \alpha \sin \beta \sin \gamma}$.

#Geometry #Proof #Wow

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
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`\sum_{i=1}^3`	$\sum_{i=1}^3$
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Comments

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 - 5 years, 8 months ago

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

Akshat Sharda - 5 years, 8 months ago

what is the area I'm supposed to prove?

Brian Wang - 5 years, 8 months ago

@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 - 5 years, 8 months ago

@Akshat Sharda – ah, ok

Brian Wang - 5 years, 8 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$