Girish, you’re right, that’s a fair bit simpler than the method I used (proved it painfully from scratch – I learnt this formula before the 1/2 * absinC formula!). However, your method too requires the inscribed angle theorem and similar triangles to prove that sinA = a/2R ðŸ™‚ So I still stand by at least part of what I said ðŸ˜‰

Sir but area of the triangle is 1/2 bc*sin a, if we draw the triangle we come to know that sin a = a/2r… so substituting we get abc / 4r

Girish, you’re right, that’s a fair bit simpler than the method I used (proved it painfully from scratch – I learnt this formula before the 1/2 * absinC formula!). However, your method too requires the inscribed angle theorem and similar triangles to prove that sinA = a/2R ðŸ™‚ So I still stand by at least part of what I said ðŸ˜‰

regards

J

i think in point 4 it should be area of triangle instead of circle

Right you are Dev, I have edited it now. Thanks.

regards

J