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 \(\frac{(k \sin \gamma + n \sin \alpha + m \sin \beta)^{2}}{2 \sin \alpha \sin \beta \sin \gamma}\).
Comments
Sort by:
Top NewestThe question is confusing. I believe you missed a number "Prove the area of a triangle with angles x,y,z knowing that" – Brian Wang · 11 months, 2 weeks ago
Log in to reply
No !! its correct. Please tell me what are you confused at. – Akshat Sharda · 11 months, 2 weeks ago
Log in to reply
what is the area I'm supposed to prove? – Brian Wang · 11 months, 2 weeks ago
Log in to reply
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 · 11 months, 2 weeks ago
Log in to reply
ah, ok – Brian Wang · 11 months, 2 weeks ago
Log in to reply