If A = 60^circ,a = 5,and b = 4sqrt{3} in Delt | vedclass

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(D) Using the Law of Sines: $\frac{a}{\sin A} = \frac{b}{\sin B}$.Rearranging for $\sin B$: $\sin B = \frac{b \sin A}{a}$.Substituting the given value

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(D) Using the Law of Sines: $\frac{a}{\sin A} = \frac{b}{\sin B}$.Rearranging for $\sin B$: $\sin B = \frac{b \sin A}{a}$.Substituting the given values: $\sin B = \frac{4\sqrt{3} \sin 60^\circ}{5} = \frac{4\sqrt{3} 	imes (\frac{\sqrt{3}}{2})}{5} = \frac{4 	imes 3}{2 	imes 5} = \frac{6}{5} = 1.2$.Since the value of $\sin B$ must be $\le 1$,the value $1.2$ is inadmissible.Therefore,no such triangle exists,and the correct option is $D$.

If $A = 60^\circ$,$a = 5$,and $b = 4\sqrt{3}$ in $\Delta ABC$,then $B =$

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