In a triangle ABC,if a^2-b^2-c^2=bc(lambda^2- | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given the equation: $a^2-b^2-c^2=bc(\lambda^2-2\lambda-1)$ Rearranging the terms,we get: $b^2+c^2-a^2 = -bc(\lambda^2-2\lambda-1)$ Dividing both s

Vedclass · Accepted Answer

$-1 \leq \lambda \leq 3$

Vedclass · Answer

(C) Given the equation: $a^2-b^2-c^2=bc(\lambda^2-2\lambda-1)$ Rearranging the terms,we get: $b^2+c^2-a^2 = -bc(\lambda^2-2\lambda-1)$ Dividing both sides by $2bc$,we have: $\frac{b^2+c^2-a^2}{2bc} = -\frac{1}{2}(\lambda^2-2\lambda-1)$ Using the Law of Cosines,$\cos A = \frac{b^2+c^2-a^2}{2bc}$,so: $\cos A = -\frac{1}{2}(\lambda^2-2\lambda-1)$ Since $-1 \leq \cos A \leq 1$,we have: $-1 \leq -\frac{1}{2}(\lambda^2-2\lambda-1) \leq 1$ Multiplying by $-2$ (and reversing the inequalities): $-2 \leq \lambda^2-2\lambda-1 \leq 2$ Adding $1$ to all parts: $-1 \leq \lambda^2-2\lambda \leq 3$ Completing the square: $-1 \leq (\lambda-1)^2-1 \leq 3$ Adding $1$ to all parts: $0 \leq (\lambda-1)^2 \leq 4$ Taking the square root: $0 \leq |\lambda-1| \leq 2$ This implies $-2 \leq \lambda-1 \leq 2$,which simplifies to: $-1 \leq \lambda \leq 3$

In a triangle $ABC$,if $a^2-b^2-c^2=bc(\lambda^2-2\lambda-1)$,then

Similar Questions

If the angles $A, B$ and $C$ of a triangle are in an Arithmetic Progression and if $a, b$ and $c$ denote the lengths of the sides opposite to $A, B$ and $C$ respectively,then the value of the expression $\frac{a}{c} \sin 2C + \frac{c}{a} \sin 2A$ is

In triangle $ABC$,if $\frac{a+b+c}{BC+AB}+\frac{a+b+c}{AC+AB}=3$,then $\tan \frac{C}{8}=$

If in $\triangle ABC$,$a \tan A + b \tan B = (a + b) \tan \left(\frac{A+B}{2}\right)$,then which of the following holds?

If $x, y, z$ are the lengths of the perpendiculars drawn from the circumcenter to the sides $a, b, c$ respectively of a triangle,then the value of $\frac{bx}{c} + \frac{cy}{a} + \frac{az}{b}$ is

If $A, B, C, D$ are angles of a cyclic quadrilateral,then $\cos A + \cos B + \cos C + \cos D$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module