In a triangle ABC,if frac{a}{b^2-c^2} + frac{ | vedclass

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

Question

(D) Given,$\frac{a}{b^2-c^2} + \frac{c}{b^2-a^2} = 0$. Using the sine rule $a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$: $\frac{2R \sin A}{4R^2

Vedclass · Accepted Answer

$\frac{\pi}{3}$

Vedclass · Answer

(D) Given,$\frac{a}{b^2-c^2} + \frac{c}{b^2-a^2} = 0$. Using the sine rule $a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$: $\frac{2R \sin A}{4R^2(\sin^2 B - \sin^2 C)} + \frac{2R \sin C}{4R^2(\sin^2 B - \sin^2 A)} = 0$ $\Rightarrow \frac{\sin A}{\sin(B+C)\sin(B-C)} + \frac{\sin C}{\sin(B+A)\sin(B-A)} = 0$ Since $A+B+C = \pi$,we have $\sin(B+C) = \sin A$ and $\sin(B+A) = \sin C$. $\Rightarrow \frac{\sin A}{\sin A \sin(B-C)} + \frac{\sin C}{\sin C \sin(B-A)} = 0$ $\Rightarrow \frac{1}{\sin(B-C)} + \frac{1}{\sin(B-A)} = 0$ $\Rightarrow \sin(B-A) + \sin(B-C) = 0$ $\Rightarrow 2 \sin\left(\frac{2B-A-C}{2}ight) \cos\left(\frac{A-C}{2}ight) = 0$ Assuming $\cos\left(\frac{A-C}{2}ight) 
eq 0$,we get $\sin\left(\frac{2B-(A+C)}{2}ight) = 0$. Since $A+C = \pi - B$,we have $\frac{2B-(\pi-B)}{2} = 0$ $\Rightarrow 3B = \pi$ $\Rightarrow B = \frac{\pi}{3}$.

In a $\triangle ABC$,if $\frac{a}{b^2-c^2} + \frac{c}{b^2-a^2} = 0$,then $B$ is equal to

Similar Questions

With the usual notation,in $\Delta ABC$,if $\angle A + \angle B = 120^{\circ}$ and $a : b = (\sqrt{3} + 1) : (\sqrt{3} - 1)$,then the ratio $\angle A : \angle B$ is

Angles of a triangle are in the ratio $4:1:1$. Then the ratio of its greatest side to its perimeter is

If in a $\Delta ABC$,$a, b, c$ are in $A.P.$,then $\tan \frac{A}{2} \tan \frac{C}{2} = $

We are given $b, c$ and $\sin B$ such that $B$ is acute and $b < c \sin B$. Then

If,in a $\triangle ABC$,$\tan \frac{A}{2} = \frac{5}{6}$ and $\tan \frac{C}{2} = \frac{2}{5}$,then $a, b, c$ are such that :

Vedclass Test Series

Exam Paper Generator

Online Exam Module