a,
+ Xét `ΔBCH` có:
`sinB=\frac{CH}{BC}`
`<=>sin60^o=\frac{CH}{12}`
`<=>CH=12.\frac{\sqrt{3}}{2}=6\sqrt{3}` (cm)
+ Xét `ΔACH` có: ``
`sin A=\frac{CH}{AC}`
`<=>sin80^o=\frac{6\sqrt{3}}{AC}`
`<=>AC=\frac{6\sqrt{3}}{sin80^o}≈10,55` (cm)
b,
+ Xét `ΔACH` có:
`AH^2=AC^2-CH^2`
`<=>AH^2=(10,55)^2-(6\sqrt{3})^2`
`<=>AH=1,82` (cm)
+ Xét `ΔCHB` có:
`HB^2=CB^2-CH^2`
`<=>HB^2=12^2-(6\sqrt{3})^2`
`<=>HB=6` (cm)
Do đó:
`S_{ΔABC}=\frac{CH.(AH+HB)}{2}=\frac{6\sqrt{3}.(1,82+6)}{2}=40,63` `(cm^2)`
Vậy `S_{ΔABC}=40,63cm^2`.
