Xét `\triangle ABC` có: đường cao `AH`
`@AB^2=BH.BC=>12^2=5.BC=>BC=28,8`
`@AB^2+AC^2=BC^2=>12^2+AC^2=28,8^2=>AC=[12\sqrt{119}]/5`
`@sin B=[AC]/[BC]=\sqrt{119}/12`
`@sin C=[AB]/[BC]=5/12`
Xét Δ \(BHA\) vuông tại H có
\(BH^2+AH^2=AB^2\\ =>AH=\sqrt{12^2-5^2}=\sqrt{119}\)
\(\odot sinB=\dfrac{AH}{AB}=\dfrac{\sqrt{119}}{12}\)
Xét \(\Delta BAC\) vuông tại A có
\(AB^2=BH.BC\\ =>BC=\dfrac{AB^2}{BH}=\dfrac{12^2}{5}=\dfrac{144}{5}\)
\(\odot sinC=\dfrac{AB}{BC}=\dfrac{12}{\dfrac{144}{5}}=\dfrac{5}{12}\)