Question

If in a triangle $ABC,$ $(a+b+c)(b+c-a)=\lambda bc,$ then :

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

Answer 1

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

Given: $(a+b+c)(b+c-a)=\lambda bc$
$\Rightarrow \left(2s\right)(a+b+c-2a)=\lambda bc$
$\Rightarrow \left(2s\right)(2s-2a)=\lambda bc$
$\Rightarrow 4\left(s\right)(s-a)=\lambda bc$
$\Rightarrow \frac{s(s-a)}{bc}=\frac{\lambda }{4}$
$\Rightarrow {\cos }^2\frac{A}{2}=\frac{\lambda }{4}$
$\Rightarrow 0<\frac{\lambda }{4}<1(\because \text{In Triangle ABC,}{\cos }^2\frac{A}{2}\in (0,1\left)\right)$
$\Rightarrow 0<\lambda <4$