In a A B C , if A=60^ , then (a+b+c)(b+c-a)=

Step-by-step solution

In A B C , A=60^ , B+ C=180^ - A=120^ a=k A, b=k B, c=k C (sine rule) Now, (b+c+a)(b+c-a) aligned & =(b+c)^2-a^2 & =k^2( B+ C)^2-k^2 ^2 A & =K^2 (2 B+C 2 B-C 2 )^2-K^2 ( 3 4 ) & [ A= 3 2 ] aligned aligned & =k^2 (2 60^ B-C 2 )^2- 3 k^2 4 & =3 k^2 ^2 B-C 2 - 3 k^2 4 & =3 k^2 ( ^2 B-C 2 - 1 4 ) & = 3 k^2 2 (2 ^2 B-C 2 -1+ 1 2 ) & = 3 k^2 2 ( (B-C)+ 1 2 ) & = 3 k^2 2 ( (B-C)+ A) & = 3 k^2 2 ( (B-C)- (B+C)) & = 3 k^2 2 2 B C aligned =3(k B)(k C)=3 b c