`a)` Xét `\triangle ABC` có:
`\qquad sin C=(AB)/(BC)=3/5`
`<=> 18/(BC)=3/5`
`<=> BC=(18.5)/3=30`
Lại có: `AB^2+AC^2=BC^2`
`<=> AC^2=BC^2-AB^2`
`=> AC^2=30^2-18^2=576`
`<=> AC=24`
Vậy `AC=24cm, BC=30cm`
`b) sin B=(AC)/(BC)=24/30=4/5`
`\qquad cosB=sin C=3/5`
`\qquad tanB=(AC)/(AB)=24/18=4/3`
`\qquad cot B=1/(tan B)=3/4`