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Câu 2. a) Ta có: $\frac{AB}{BC}=\sin B=\frac{2}{3}$ Suy ra: $AB=\frac{2}{3}\times BC=\frac{2}{3}\times 6=4(cm)$ b) Ta có: $\frac{AB}{BC}=\sin B=\frac{2}{3}$ Suy ra: $BC=\frac{3}{2}\times AB=\frac{3}{2}\times 12=18(cm)$ c) Ta có: $\widehat{A}=90^0$ nên $\widehat{B}+\widehat{C}=90^0$ Mà $\widehat{B}=53^08^\prime$ nên $\widehat{C}=90^0-53^08^\prime=36^052^\prime$ d) Ta có: $\widehat{A}=90^0$ nên $\widehat{B}+\widehat{C}=90^0$ Mà $\widehat{B}=53^08^\prime$ nên $\widehat{C}=90^0-53^08^\prime=36^052^\prime$ $\sin C=\sin 36^052^\prime\ne \frac{2}{3}$