In a triangle ABC,if (a+b+c)(b+c-a) = lambda | vedclass

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(C) Given the expression $(a+b+c)(b+c-a) = \lambda bc$. This can be rewritten as $((b+c)+a)((b+c)-a) = \lambda bc$. Using the identity $(x+y)(x-y) = x

Vedclass · Accepted Answer

$0 < \lambda < 4$

Vedclass · Answer

(C) Given the expression $(a+b+c)(b+c-a) = \lambda bc$. This can be rewritten as $((b+c)+a)((b+c)-a) = \lambda bc$. Using the identity $(x+y)(x-y) = x^2 - y^2$,we get $(b+c)^2 - a^2 = \lambda bc$. Expanding this,$b^2 + c^2 + 2bc - a^2 = \lambda bc$. Rearranging the terms,$b^2 + c^2 - a^2 = (\lambda - 2)bc$. Dividing both sides by $2bc$,we get $\frac{b^2 + c^2 - a^2}{2bc} = \frac{\lambda - 2}{2}$. By the Law of Cosines,$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$,so $\cos A = \frac{\lambda - 2}{2}$. Since $-1 Multiplying by $2$,$-2 Adding $2$ to all parts,$0 < \lambda < 4$.

In a $\triangle ABC$,if $(a+b+c)(b+c-a) = \lambda bc$,then which of the following is true?

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