a=|AB|=3\sqrt2
b=6
\beta=45^\circ
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\frac{b}{\sin\beta}=\frac{6}{sin45^{\circ}}=\frac{6}{\frac{\sqrt2}{2}}=6*\frac{2}{\sqrt2}=\frac{12*\sqrt2}{\sqrt2*\sqrt2}=\frac{12\sqrt2}{2}=6\sqrt2
z twierdzenia sinusów
\frac{a}{\sin\alpha}=\frac{b}{\sin\beta}
\frac{3\sqrt2}{\sin\alpha}=6\sqrt2 \ |:3\sqrt2
\frac{1}{\sin\alpha}=2
2\sin\alpha=1 \ |:2
sin\alpha=\frac{1}{2}
\alpha=30^\circ …150^\circ odrzucam 150+45>180
\gamma=180^\circ-(45^\circ+30^\circ)=105^\circ
Odpowiedź:
\alpha=30^\circ , \ \beta=45^\circ , \ \gamma= 105^\circ