If A and B are acute angles satisfying 3 cos | 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}$ $\Rightarrow 3 \sin A \cos A = 2 \s

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(D) Given equations are:$1) \ 3 \cos^2 A + 2 \cos^2 B = 4$$2) \ \frac{3 \sin A}{\sin B} = \frac{2 \cos B}{\cos A}$ $\Rightarrow 3 \sin A \cos A = 2 \sin B \cos B$ $\Rightarrow \frac{3}{2} \sin 2A = \sin 2B$From $(1)$:$3(1 - \sin^2 A) + 2(1 - \sin^2 B) = 4$$5 - 3 \sin^2 A - 2 \sin^2 B = 4$$3 \sin^2 A + 2 \sin^2 B = 1$$3 \sin^2 A = 1 - 2 \sin^2 B = \cos 2B$Consider $\cos(A + 2B) = \cos A \cos 2B - \sin A \sin 2B$Substitute $\cos 2B = 3 \sin^2 A$ and $\sin 2B = \frac{3}{2} \sin 2A = 3 \sin A \cos A$:$\cos(A + 2B) = \cos A (3 \sin^2 A) - \sin A (3 \sin A \cos A)$$\cos(A + 2B) = 3 \sin^2 A \cos A - 3 \sin^2 A \cos A = 0$Since $A, B$ are acute,$A + 2B$ must be $90^{\circ}$.

If $A$ and $B$ are 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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