In triangle ABC,if sin^2 B = sin C and 3 cos^ | vedclass

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(D) Given $3 \cos^2 B = 2 \cos^2 C$. Substituting $\cos^2 	heta = 1 - \sin^2 	heta$,we get $3(1 - \sin^2 B) = 2(1 - \sin^2 C)$. Since $\sin^2 B = \s

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a scalene triangle

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(D) Given $3 \cos^2 B = 2 \cos^2 C$. Substituting $\cos^2 	heta = 1 - \sin^2 	heta$,we get $3(1 - \sin^2 B) = 2(1 - \sin^2 C)$. Since $\sin^2 B = \sin C$,we substitute this into the equation: $3(1 - \sin C) = 2(1 - \sin^2 C)$. $3 - 3 \sin C = 2 - 2 \sin^2 C$. $2 \sin^2 C - 3 \sin C + 1 = 0$. $(2 \sin C - 1)(\sin C - 1) = 0$. Thus,$\sin C = \frac{1}{2}$ or $\sin C = 1$. If $\sin C = 1$,then $C = \frac{\pi}{2}$,which implies $\sin^2 B = \sin C = 1$,so $B = \frac{\pi}{2}$. This is impossible in a triangle as $B+C = \pi$. Therefore,$\sin C = \frac{1}{2}$,so $C = \frac{\pi}{6}$ or $C = \frac{5\pi}{6}$. If $C = \frac{\pi}{6}$,then $\sin^2 B = \sin(\frac{\pi}{6}) = \frac{1}{2}$,so $\sin B = \frac{1}{\sqrt{2}}$,which means $B = \frac{\pi}{4}$ or $B = \frac{3\pi}{4}$. If $B = \frac{\pi}{4}$ and $C = \frac{\pi}{6}$,then $A = \pi - (\frac{\pi}{4} + \frac{\pi}{6}) = \pi - \frac{5\pi}{12} = \frac{7\pi}{12}$. Since all angles $A, B, C$ are distinct,$	riangle ABC$ is a scalene triangle.

In $\triangle ABC$,if $\sin^2 B = \sin C$ and $3 \cos^2 B = 2 \cos^2 C$,then $\triangle ABC$ is

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