If a = 2 sqrt{2},b = 6,and A = 45^{circ},then | vedclass

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(A) Given: $a = 2 \sqrt{2}$,$b = 6$,and $A = 45^{\circ}$.Using the Law of Sines: $\frac{a}{\sin A} = \frac{b}{\sin B}$.Substituting the values: $\frac

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no triangle is possible

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(A) Given: $a = 2 \sqrt{2}$,$b = 6$,and $A = 45^{\circ}$.Using the Law of Sines: $\frac{a}{\sin A} = \frac{b}{\sin B}$.Substituting the values: $\frac{2 \sqrt{2}}{\sin 45^{\circ}} = \frac{6}{\sin B}$.$\sin B = \frac{b \sin A}{a} = \frac{6 	imes \sin 45^{\circ}}{2 \sqrt{2}}$.$\sin B = \frac{6 	imes \frac{1}{\sqrt{2}}}{2 \sqrt{2}} = \frac{6}{2 	imes 2} = \frac{6}{4} = 1.5$.Since the value of $\sin B$ cannot exceed $1$,$\sin B = 1.5$ is impossible.Therefore,no triangle is possible.

If $a = 2 \sqrt{2}$,$b = 6$,and $A = 45^{\circ}$,then:

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