Question

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

If$(a+b+c)(b+c-a)=\lambda bc$, then ?

• A. $\lambda <0$
• B. $\lambda >6$
• C. $0<\lambda <4$
• D. $\lambda >4$

Answer 1

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

Given that $(a+b+c)(b+c-a)=\lambda bc$
$\Rightarrow \lambda =\frac{{(b+c)}^2-a^2}{bc}=\frac{b^2+c^2+2bc-a^2}{bc}=\frac{b^2+c^2-a^2}{bc}+2=2\cos A+2$
.
$\Rightarrow \lambda =4{\cos }^2\frac{A}{2}$
.
Since $0<{\cos }^2\frac{A}{2}<1$ in a triangle, we get $0<\lambda <4$