Question

In 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 $0<\lambda <4$

$\begin{array}{c}=(a+(b+c\left)\right)\left(\right(b+c)-a)\\ =(\frac{6+c{)}^2-a^2}{b^2+c^2-a^2+2bc}=\lambda -2\\ \Rightarrow b^2+c^2-a^2+2bc=\lambda bc\\ \Rightarrow \frac{b^2+c^2-a^2}{bc}=\lambda -2\\ \Rightarrow \frac{b^2+c^2-a^2}{2bc}=\cos A=\frac{\lambda -2}{2}\\ -1<\cos A<1\\ \Rightarrow -1<\frac{\lambda -2}{2}<1\\ \Rightarrow -2<\lambda -2<2\\ \Rightarrow 0<\lambda <4\\ \text{Option}C\text{is correct}\end{array}$