Let alpha = 3 sin^{-1}(frac{6}{11}) and beta | vedclass

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Question

(A) Given $\alpha = 3 \sin^{-1}(\frac{6}{11})$. Since $\frac{6}{11} > \frac{1}{2}$,we have $\sin^{-1}(\frac{6}{11}) > \sin^{-1}(\frac{1}{2}) = 30^\cir

Vedclass · Accepted Answer

Both Statement $I$ and Statement $II$ are true

Vedclass · Answer

(A) Given $\alpha = 3 \sin^{-1}(\frac{6}{11})$. Since $\frac{6}{11} > \frac{1}{2}$,we have $\sin^{-1}(\frac{6}{11}) > \sin^{-1}(\frac{1}{2}) = 30^\circ$. Thus,$\alpha > 3 	imes 30^\circ = 90^\circ$. Since $90^\circ Given $\beta = 3 \cos^{-1}(\frac{4}{9})$. Since $\frac{4}{9}  \cos^{-1}(\frac{1}{2}) = 60^\circ$. Thus,$\beta > 3 	imes 60^\circ = 180^\circ$. Since $\alpha > 90^\circ$ and $\beta > 180^\circ$,$\alpha + \beta > 270^\circ$. In the fourth quadrant,the cosine function is positive. Therefore,$\cos(\alpha + \beta) > 0$. So,Statement $I$ is true.

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