In a triangle ABC,if sin^2 B = sin A and 2 co | vedclass

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(B) Given: $\sin^2 B = \sin A$ and $2 \cos^2 A = 3 \cos^2 B$. Using $\cos^2 	heta = 1 - \sin^2 	heta$,we have $2(1 - \sin^2 A) = 3(1 - \sin^2 B)$. S

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obtuse angled

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(B) Given: $\sin^2 B = \sin A$ and $2 \cos^2 A = 3 \cos^2 B$. Using $\cos^2 	heta = 1 - \sin^2 	heta$,we have $2(1 - \sin^2 A) = 3(1 - \sin^2 B)$. Substituting $\sin^2 B = \sin A$,we get $2 - 2 \sin^2 A = 3 - 3 \sin A$. Rearranging gives $2 \sin^2 A - 3 \sin A + 1 = 0$. Factoring the quadratic equation: $(2 \sin A - 1)(\sin A - 1) = 0$. This implies $\sin A = 1/2$ or $\sin A = 1$. If $\sin A = 1$,then $A = 90^\circ$. If $A = 90^\circ$,then $\sin^2 B = \sin 90^\circ = 1$,so $B = 90^\circ$. This is impossible in a triangle as $A+B+C = 180^\circ$. If $\sin A = 1/2$,then $A = 30^\circ$ or $A = 150^\circ$. If $A = 30^\circ$,then $\sin^2 B = \sin 30^\circ = 1/2$,so $\cos^2 B = 1 - 1/2 = 1/2$. Checking the second equation: $2 \cos^2 A = 2 \cos^2 30^\circ = 2(3/4) = 3/2$. $3 \cos^2 B = 3(1/2) = 3/2$. Since $3/2 = 3/2$,the condition holds. Since $A = 30^\circ$,$B = 45^\circ$ or $135^\circ$. If $B = 45^\circ$,$C = 180^\circ - 30^\circ - 45^\circ = 105^\circ$ (obtuse). If $B = 135^\circ$,$A+B = 165^\circ In both cases,the triangle contains an obtuse angle.

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

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