Question

$ln$ a triangle $ABC,(a+b+c)(b+c-a)=\lambda bc$ if

• A. $\lambda <0$
• B. $\lambda >0$
• C. $0<\lambda <4$
• D. $\lambda >4$

Answer 1

The correct option is C $0<\lambda <4$

$(b+c{)}^2-a^2=\lambda bc$
$b^2+c^2+2bc-a^2=\lambda bc$
$b^2+c^2-a^2=(\lambda -2)bc$
$\lambda -2=\frac{b^2+c^2-a^2}{bc}$
$\cos A=\frac{\lambda -2}{2}$
$-1<\frac{\lambda -2}{2}<1$
$-2<\lambda -2<2$
$0<\lambda <4$