Xét `\triangle ABC` vuông tại `A` có:
`@\hat{C}=90^o -\hat{B}=90^o -50^o =40^o`
`@sin C=[AB]/[BC]<=>sin 40^o =[AB]/20<=>AB~~12,86(cm)`
`@cos C=[AC]/[BC]<=>cos 40^o =[AC]/20<=>AC~~15,32(cm)`
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Xét `\triangle ABC` vuông tại `A` có:
`@AB^2+AC^2=BC^2<=>18^2+21^2=BC^2<=>BC=3\sqrt{85}(cm)`
`@sin B=[AC]/[BC]=21/[3\sqrt{85}]=>\hat{B}~~49^o`
`@\hat{C}=90^o -\hat{B}~~90^o -49^o ~~41^o`