Question

In a $\Delta ABC,\left( a+b+c \right)\left( b+c-a \right)=\lambda bc$, then
a.$\lambda <-6$
b.$\lambda >6$
c.$0<\lambda <4$
d.$\lambda >4$

Answer 1

Hint: Apply basic trigonometric property and simplify the given expression. Then use the law of cosines in a triangle and get an equation with cosine function. Substitute and this takes the range of cosine function as $-1<\cos A<1$.

Complete step-by-step answer:

Given to us a triangle ABC, let us consider the three sides of the triangle as a, b, c.
It is given that, $\left( a+b+c \right)\left( b+c-a \right)=\lambda bc$.
We can arrange the terms and write as,
$\left( \left( b+c \right)+a \right)\left( \left( b+c \right)-a \right)=\lambda bc-(1)$
Now this is of the form, $\left( x+y \right)\left( x-y \right)={{x}^{2}}-{{y}^{2}}$.
$\therefore {{\left( b+c \right)}^{2}}-{{a}^{2}}=\lambda bc$
Now, ${{\left( x+y \right)}^{2}}={{x}^{2}}+{{y}^{2}}+2xy$. Let us expand, ${{\left( b+c \right)}^{2}}$.

& \therefore {{b}^{2}}+{{c}^{2}}+2bc-{{a}^{2}}=\lambda bc \\\ & {{b}^{2}}+{{c}^{2}}-{{a}^{2}}=\lambda bc \\\ & {{b}^{2}}+{{c}^{2}}-{{a}^{2}}=\left( \lambda -2 \right)bc-(2) \\\ \end{aligned}$$ From the triangle, a, b and c are sides.  A is the angle opposite to side a. By law of cosines, we can say that, $$\begin{aligned} & {{a}^{2}}={{b}^{2}}+{{c}^{2}}-2bc\cos A \\\ & \therefore {{b}^{2}}+{{c}^{2}}-{{a}^{2}}=2bc\cos A-(3) \\\ \end{aligned}$$ Now substitute equation (3) in equation (2). $$\begin{aligned} & \therefore {{b}^{2}}+{{c}^{2}}-{{a}^{2}}=\left( \lambda -2 \right)bc \\\ & 2bc\cos A=\left( \lambda -2 \right)bc \\\ \end{aligned}$$ Cancel bc from LHS and RHS. $$\therefore \cos A=\dfrac{\lambda -2}{2}$$ Now for $$\cos A$$, $$-1<\cos A<1$$. $$\therefore $$ Put, $$\cos A=\dfrac{\lambda -2}{2}$$ $$\begin{aligned} & \Rightarrow -1<\dfrac{\lambda -2}{2}<1 \\\ & =2<\lambda -2<2 \\\ & =-2+2<\lambda <2+2 \\\ & \therefore 0<\lambda <4 \\\ \end{aligned}$$ Thus we got the range of $$\lambda $$ as $$0<\lambda <4$$. $$\therefore $$ Option (c) is the correct answer. Note: The Law of cosine is useful for finding the $${{3}^{rd}}$$ side of a triangle when we know 2 sides and angle between them. And it is used for finding the angles of a triangle when we know all three sides.