If in a triangle ABC,a = 5,b = 4,and A = frac | vedclass

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(D) Given $\frac{a}{\sin A} = \frac{b}{\sin B}$.Substituting $a=5, b=4$ and $A = \frac{\pi}{2} + B$:$\frac{5}{\sin(\frac{\pi}{2} + B)} = \frac{4}{\sin

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$2	an^{-1}\left(\frac{1}{9}ight)$

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(D) Given $\frac{a}{\sin A} = \frac{b}{\sin B}$.Substituting $a=5, b=4$ and $A = \frac{\pi}{2} + B$:$\frac{5}{\sin(\frac{\pi}{2} + B)} = \frac{4}{\sin B} \implies \frac{5}{\cos B} = \frac{4}{\sin B}$.Thus,$	an B = \frac{4}{5}$.Since $A = \frac{\pi}{2} + B$,$	an A = 	an(\frac{\pi}{2} + B) = -\cot B = -\frac{5}{4}$.In $	riangle ABC$,$C = \pi - (A + B)$.$	an C = 	an(\pi - (A + B)) = -	an(A + B) = -\frac{	an A + 	an B}{1 - 	an A 	an B}$.Substituting values: $	an C = -\frac{-\frac{5}{4} + \frac{4}{5}}{1 - (-\frac{5}{4})(\frac{4}{5})} = -\frac{-\frac{25}{20} + \frac{16}{20}}{1 + 1} = -\frac{-\frac{9}{20}}{2} = \frac{9}{40}$.Using the identity $2	an^{-1}(x) = 	an^{-1}(\frac{2x}{1-x^2})$,we check $2	an^{-1}(\frac{1}{9}) = 	an^{-1}(\frac{2/9}{1 - 1/81}) = 	an^{-1}(\frac{2/9}{80/81}) = 	an^{-1}(\frac{2}{9} 	imes \frac{81}{80}) = 	an^{-1}(\frac{9}{40})$.Therefore,$C = 2	an^{-1}(\frac{1}{9})$.

If in a triangle $ABC$,$a = 5$,$b = 4$,and $A = \frac{\pi}{2} + B$,then $C$ is:

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