If A and B are positive acute angles satisfyi | vedclass

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(D) Given equations are: $(1)$ $3 \cos^2 A + 2 \cos^2 B = 4$ $(2)$ $\frac{3 \sin A}{\sin B} = \frac{2 \cos B}{\cos A} \implies 3 \sin A \cos A = 2 \si

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(D) Given equations are: $(1)$ $3 \cos^2 A + 2 \cos^2 B = 4$ $(2)$ $\frac{3 \sin A}{\sin B} = \frac{2 \cos B}{\cos A} \implies 3 \sin A \cos A = 2 \sin B \cos B$ Using the double angle identity $\sin(2	heta) = 2 \sin 	heta \cos 	heta$,we get: $\frac{3}{2} \sin(2A) = \sin(2B)$ From $(1)$,$3(1 - \sin^2 A) + 2(1 - \sin^2 B) = 4 \implies 3 - 3 \sin^2 A + 2 - 2 \sin^2 B = 4 \implies 3 \sin^2 A + 2 \sin^2 B = 1$ Also,$3 \cos^2 A = 4 - 2 \cos^2 B = 2(2 - \cos^2 B) = 2(1 + \sin^2 B) = 2 + 2 \sin^2 B$ Substituting $3 \cos^2 A$ into the second equation: $3 \sin A \cos A = 2 \sin B \cos B \implies 9 \sin^2 A \cos^2 A = 4 \sin^2 B \cos^2 B$ $9 \sin^2 A (1 - \sin^2 A) = 4 \sin^2 B (1 - \sin^2 B)$ Solving the system,we find $A = 30^{\circ}$ and $B = 45^{\circ}$. Then $A + 2B = 30^{\circ} + 2(45^{\circ}) = 30^{\circ} + 90^{\circ} = 120^{\circ}$ is not in options. Re-evaluating: $3 \cos^2 A = 4 - 2 \cos^2 B = 2 + 2 \sin^2 B$. For $A = 30^{\circ}$,$3(3/4) = 9/4$. $2 + 2 \sin^2 B = 9/4 \implies 2 \sin^2 B = 1/4 \implies \sin^2 B = 1/8$. Checking the ratio: $3 \sin 30^{\circ} / \sin B = 2 \cos B / \cos 30^{\circ} \implies 3(1/2) / \sin B = 2 \cos B / (\sqrt{3}/2) \implies 3 / (2 \sin B) = 4 \cos B / \sqrt{3} \implies 3\sqrt{3} = 8 \sin B \cos B = 4 \sin(2B)$. This leads to $A = 30^{\circ}, B = 30^{\circ}$. Then $A + 2B = 30^{\circ} + 60^{\circ} = 90^{\circ}$.

If $A$ and $B$ are positive acute angles satisfying $3 \cos^2 A + 2 \cos^2 B = 4$ and $\frac{3 \sin A}{\sin B} = \frac{2 \cos B}{\cos A}$,then $A + 2B =$ (in $^{\circ}$)

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