Xét `ΔABC` vuông tại `A`
`1/(AH)^2=1/(AB)^2+1/(AC)^2`
`⇔1/(24^2)=1/(30^2)+1/(AC)^2`
`⇔1/1600=1/(AC)^2`
`⇒AC=40`
`AB^2+AC^2=BC^2`
`⇔BC=\sqrt{30^2+40^2}=50`
`⇒sin B=(AC)/(BC)=40/50=4/5`
`⇒cos B=(AB)/(BC)=30/50=3/5`
`⇒tan B=(AC)/(AB)=40/30=4/3`
`⇒cot B=(AB)/(AC)=30/40=3/4`