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, \[{{\left( b+c \right)}^{2}}\].
\[\begin{align}
  & \therefore {{b}^{2}}+{{c}^{2}}+2bc-{{a}^{2}}=\lambda bc \\
 & {{b}^{2}}+{{c}^{2}}-{{a}^{2}}=\lambda bc \\
 & {{b}^{2}}+{{c}^{2}}-{{a}^{2}}=\left( \lambda -2 \right)bc-(2) \\
\end{align}\]
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,
\[\begin{align}
  & {{a}^{2}}={{b}^{2}}+{{c}^{2}}-2bc\cos A \\
 & \therefore {{b}^{2}}+{{c}^{2}}-{{a}^{2}}=2bc\cos A-(3) \\
\end{align}\]
Now substitute equation (3) in equation (2).
\[\begin{align}
  & \therefore {{b}^{2}}+{{c}^{2}}-{{a}^{2}}=\left( \lambda -2 \right)bc \\
 & 2bc\cos A=\left( \lambda -2 \right)bc \\
\end{align}\]
Cancel bc from LHS and RHS.
\[\therefore \cos A=\dfrac{\lambda -2}{2}\]
Now for \[\cos A\], \[-1<\cos A<1\].
\[\therefore \] Put, \[\cos A=\dfrac{\lambda -2}{2}\]
\[\begin{align}
  & \Rightarrow -1<\dfrac{\lambda -2}{2}<1 \\
 & =2<\lambda -2<2 \\
 & =-2+2<\lambda <2+2 \\
 & \therefore 0<\lambda <4 \\
\end{align}\]
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.