If A and B are acute angles such that sin A = | vedclass

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(A) Given: $\sin A = \sin^2 B$ and $2 \cos^2 A = 3 \cos^2 B$. Since $A$ and $B$ are acute, $\sin A, \sin B, \cos A, \cos B > 0$. Substitute $\cos^2 A

Vedclass · Accepted Answer

$\left(\frac{\pi}{6}, \frac{\pi}{4}\right)$

Vedclass · Answer

(A) Given: $\sin A = \sin^2 B$ and $2 \cos^2 A = 3 \cos^2 B$. Since $A$ and $B$ are acute, $\sin A, \sin B, \cos A, \cos B > 0$. Substitute $\cos^2 A = 1 - \sin^2 A$ and $\cos^2 B = 1 - \sin^2 B$ into the second equation: $2(1 - \sin^2 A) = 3(1 - \sin^2 B)$. Substitute $\sin^2 B = \sin A$: $2 - 2 \sin^2 A = 3(1 - \sin A) = 3 - 3 \sin A$. $2 \sin^2 A - 3 \sin A + 1 = 0$. $(2 \sin A - 1)(\sin A - 1) = 0$. So, $\sin A = \frac{1}{2}$ or $\sin A = 1$. Since $A$ is an acute angle, $\sin A = 1$ implies $A = \frac{\pi}{2}$, which is not acute. Thus, $\sin A = \frac{1}{2}$, which gives $A = \frac{\pi}{6}$. Then $\sin^2 B = \sin A = \frac{1}{2}$, so $\sin B = \frac{1}{\sqrt{2}}$ (since $B$ is acute). Therefore, $B = \frac{\pi}{4}$. The solution is $(A, B) = \left(\frac{\pi}{6}, \frac{\pi}{4}\right)$.

If $A$ and $B$ are acute angles such that $\sin A = \sin^2 B$ and $2 \cos^2 A = 3 \cos^2 B$, then $(A, B) =$

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