Re: Proof using Law of Sines

I Think in a triangle, A + B + C = 180 degrees. Therefore, A = 180 - (B + C). Subtituting in the equation Sin A = Sin B Cos C + Sin C cos B, we get Sin [180 - (B + C)] = Sin B Cos C + Sin C cos B. Since in the Second Quadrant, Sine is positive and Sine remains Sine, Sin (B + C) = Sin B Cos C + Sin C cos B. Hence proved. I do not know how to express mathematically. Hope it is right. Good Souls comment on this.

Re: Proof using Law of Sines

Sin A = Sin (180 - [B + C]) = Sin (180 - B - C) = Sin (B + C)

Does this sound right?

Re: Proof using Law of Sines

Quote:

Originally Posted by

**scruz10** Sin A = Sin (180 - [B + C]) = Sin (180 - B - C) = Sin (B + C)

Does this sound right?

s(Evilgrin)