A circle inscribed in a triangle, the radius of the circle is $r$, prove that the area of triangle is $2r(\sin A+\sin B+\sin C).$
P.s i can't seem to upload the image
2026-04-03 22:08:49.1775254129
Prove that the area of a triangle with inradius $r$ equals $2r(\sin A+\sin B+\sin C)$
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This is clearly wrong by dimensional analysis. By the sine rule, your expectation is $2r(a+b+c)/(2R)=2rs/R$, with $s$ the semiperimeter and $R$ the circumradius. The area is actually $rs$. Perhaps what you really wanted to prove is that $2R(\sin A+\cdots)$ is the perimeter, which follows as above.