If alpha, beta are acute angles such that fra | vedclass

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(A) Given: $\sin \alpha = \frac{6}{5} \sin \beta$ and $\cos \alpha = \frac{9}{5 \sqrt{5}} \cos \beta$. Using the identity $\sin^2 \alpha + \cos^2 \alp

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$\frac{4}{5}$

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(A) Given: $\sin \alpha = \frac{6}{5} \sin \beta$ and $\cos \alpha = \frac{9}{5 \sqrt{5}} \cos \beta$. Using the identity $\sin^2 \alpha + \cos^2 \alpha = 1$: $(\frac{6}{5} \sin \beta)^2 + (\frac{9}{5 \sqrt{5}} \cos \beta)^2 = 1$ $\frac{36}{25} \sin^2 \beta + \frac{81}{125} \cos^2 \beta = 1$ Multiply by $125$: $180 \sin^2 \beta + 81 \cos^2 \beta = 125$ Since $\cos^2 \beta = 1 - \sin^2 \beta$: $180 \sin^2 \beta + 81(1 - \sin^2 \beta) = 125$ $180 \sin^2 \beta + 81 - 81 \sin^2 \beta = 125$ $99 \sin^2 \beta = 44$ $\sin^2 \beta = \frac{44}{99} = \frac{4}{9}$ $\sin \beta = \frac{2}{3}$ (since $\beta$ is acute). Now,$\sin \alpha = \frac{6}{5} \sin \beta = \frac{6}{5} 	imes \frac{2}{3} = \frac{4}{5}$.

If $\alpha, \beta$ are acute angles such that $\frac{\sin \alpha}{\sin \beta} = \frac{6}{5}$ and $\frac{\cos \alpha}{\cos \beta} = \frac{9}{5 \sqrt{5}}$,then $\sin \alpha = $

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