If in a triangle sin A : sin C = sin (A - B) | vedclass

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(A) Given $\frac{\sin A}{\sin C} = \frac{\sin (A - B)}{\sin (B - C)}$.Using the sine rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R

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are in $A.P.$

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(A) Given $\frac{\sin A}{\sin C} = \frac{\sin (A - B)}{\sin (B - C)}$.Using 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 $\sin C = \frac{c}{2R}$.In any triangle,$A + B + C = \pi$,so $A = \pi - (B + C)$,which implies $\sin A = \sin(B + C)$. Similarly,$C = \pi - (A + B)$,so $\sin C = \sin(A + B)$.Substituting these into the given equation:$\frac{\sin(B + C)}{\sin(A + B)} = \frac{\sin(A - B)}{\sin(B - C)}$$\sin(B + C) \sin(B - C) = \sin(A + B) \sin(A - B)$Using the identity $\sin(x+y)\sin(x-y) = \sin^2 x - \sin^2 y$:$\sin^2 B - \sin^2 C = \sin^2 A - \sin^2 B$$2 \sin^2 B = \sin^2 A + \sin^2 C$Multiplying by $(2R)^2$:$2b^2 = a^2 + c^2$Therefore,$a^2, b^2, c^2$ are in $A.P.$

If in a triangle $\sin A : \sin C = \sin (A - B) : \sin (B - C)$,then $a^2, b^2, c^2$:

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