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If $\mathrm{a}=2 \sqrt{2}, \mathrm{~b}=6, \mathrm{~A}=45^{\circ}$, then
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Verified Answer
The correct answer is:
no triangle is possible
Hints: $\mathrm{a}=2 \sqrt{2} ; \mathrm{b}=6 ; \mathrm{A}=45^0$
$$
\begin{aligned}
& \frac{\mathrm{a}}{\sin \mathrm{A}}=\frac{\mathrm{b}}{\sin \mathrm{B}} \Rightarrow \sin \mathrm{B}=\frac{\mathrm{b}}{\mathrm{a}} \sin \mathrm{A} \\
& \Rightarrow \sin \mathrm{B}=\frac{6}{2 \sqrt{2}} \sin 45^{\circ}=\frac{3}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}=\frac{3}{2} \Rightarrow \text { No triangle is possible since } \sin \mathrm{B}>1
\end{aligned}
$$
$$
\begin{aligned}
& \frac{\mathrm{a}}{\sin \mathrm{A}}=\frac{\mathrm{b}}{\sin \mathrm{B}} \Rightarrow \sin \mathrm{B}=\frac{\mathrm{b}}{\mathrm{a}} \sin \mathrm{A} \\
& \Rightarrow \sin \mathrm{B}=\frac{6}{2 \sqrt{2}} \sin 45^{\circ}=\frac{3}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}=\frac{3}{2} \Rightarrow \text { No triangle is possible since } \sin \mathrm{B}>1
\end{aligned}
$$
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