Question

In $\Delta$ $ABC$ the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively.

In a triangle ABC if $(a+b+c)(b+c-a)=xbc$, then $x$ can be equal to

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $1$
B $2$
C $3$

$(a+b+c)(b+c-a)=xbcab+ac-a^2+b^2+bc-ab+bc+c^2-ac=xbc{b}^2+c^2-a^2=(x-2)bc\frac{b^2+c^2-a^2}{2bc}=\frac{x-2}{2}\frac{x-2}{2}=\cos A$

$-1<\frac{x-2}{2}<1\frac{x-2}{2}<1x-2<2x<4$

1,2,3.

(A),(B),(C) are correct.