Xét $∆ACH$ vuông tại $H$
`=>AC^2=AH^2+CH^2` (định lý Pytago)
`=>CH^2=AC^2-AH^2=10^2-6^2=64`
`=>CH=\sqrt{64}=8cm`
$\\$
`\qquad sinC= {AH}/{AC}=6/{10}=3/5`
`\qquad cosC={CH}/{AC}=8/{10}=4/5`
`\qquad tanC={AH}/{CH}=6/8=3/4`
`\qquad cotC={CH}/{AH}=8/6=4/3`
$\\$
Xét $∆ABC$ vuông tại $A$
`=>\hat{B}+\hat{C}=90°` (hai góc phụ nhau)
`=>sinB=cosC=4/5`
`\qquad cosB=sinC=3/5`
`\qquad tanB=cotC=4/3`
`\qquad cotB=tanC=3/4`
Vậy:
`sinC=3/5;cosC=4/5;tanC=3/4;cotC=4/3`
`sinB=4/5;cosB=3/5;tanB=4/3;cotB=3/4`
