[tex] \displaystyle\\
\text{Vezi desenul atasat.}\\\\
\text{Aplicam teorema sinusurilor:}\\\\
\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\\\\
\sin B=\frac{b\sin C}{c}= \frac{\sqrt{2} \cdot \dfrac{\sqrt{3}}{2}}{\sqrt{3}}= \frac{\sqrt{2} \cdot \sqrt{3}}{2\sqrt{3}}=\frac{\sqrt{2}}{2}\\\\
\Longrightarrow~~B = 45^o\\\\
A = 180-60-45 = 75^o [/tex]
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[tex] \displaystyle\\
\sin 75 = \sin(30+45)=\sin 30\cos45+\cos 30 \sin 45 =\\\\
=\frac{1}{2}\cdot \frac{\sqrt{2}}{2}+\frac{\sqrt{3}}{2}\cdot \frac{\sqrt{2}}{2} =\frac{\sqrt{2}}{4}+ \frac{\sqrt{3}\cdot \sqrt{2}}{4}=\boxed{\frac{\sqrt{2}+\sqrt{6}}{4}}\\\\
\text{Rescriem teorema sinusurilor:} [/tex]
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[tex] \displaystyle\\
\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\\\\
a=\frac{b\sin A}{\sin B}=\frac{\sqrt{2}\cdot \dfrac{\sqrt{2}+\sqrt{6}}{4}}{\dfrac{\sqrt{2}}{2}}=\\\\
=\sqrt{2}\cdot \dfrac{\sqrt{2}+\sqrt{6}}{4}\cdot \dfrac{2}{\sqrt{2}}= \boxed{\bf \frac{\sqrt{2}+\sqrt{6}}{2}~cm } \\\\
\text{Concluzie:}\\
A= 75^o\\
B=45^o\\
C=60^o\\\\
a = \frac{\sqrt{2}+\sqrt{6}}{2}~cm\\\\
b= \sqrt{2}~cm\\
c= \sqrt{3}~cm\\ [/tex]
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