Let frac{sin A}{sin B} = frac{sin (A-C)}{sin | vedclass

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(B) Given $\frac{\sin A}{\sin B} = \frac{\sin (A-C)}{\sin (C-B)}$.Since $A, B, C$ are angles of a triangle,$A+B+C = \pi$,so $A = \pi - (B+C)$.Thus,$\s

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$b^{2}, c^{2}, a^{2}$ are in $A.P.$

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(B) Given $\frac{\sin A}{\sin B} = \frac{\sin (A-C)}{\sin (C-B)}$.Since $A, B, C$ are angles of a triangle,$A+B+C = \pi$,so $A = \pi - (B+C)$.Thus,$\sin A = \sin (B+C)$.Also,$B = \pi - (A+C)$,so $\sin B = \sin (A+C)$.Substituting these into the given equation:$\frac{\sin (B+C)}{\sin (A+C)} = \frac{\sin (A-C)}{\sin (C-B)}$.Cross-multiplying gives:$\sin (B+C) \sin (C-B) = \sin (A+C) \sin (A-C)$.Using the identity $\sin (x+y) \sin (x-y) = \sin^{2} x - \sin^{2} y$:$\sin^{2} C - \sin^{2} B = \sin^{2} A - \sin^{2} C$.Rearranging terms:$2 \sin^{2} C = \sin^{2} A + \sin^{2} B$.By the sine rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$,so $\sin A = \frac{a}{2R}$,$\sin B = \frac{b}{2R}$,and $\sin C = \frac{c}{2R}$.Substituting these into the equation:$2 \left(\frac{c}{2R}\right)^{2} = \left(\frac{a}{2R}\right)^{2} + \left(\frac{b}{2R}\right)^{2}$.$2c^{2} = a^{2} + b^{2}$.This implies $a^{2}, c^{2}, b^{2}$ are in $A.P.$ (or $b^{2}, c^{2}, a^{2}$ are in $A.P.$).

Let $\frac{\sin A}{\sin B} = \frac{\sin (A-C)}{\sin (C-B)}$,where $A, B, C$ are angles of a triangle $ABC$. If the lengths of the sides opposite these angles are $a, b, c$ respectively,then:

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