Ta có:

- $\sin B = \dfrac{3}{4} = \dfrac{AH}{AB} \Rightarrow AH = \dfrac{3}{4} \cdot 8 = 6$ (cm)
- $BH = \sqrt{AB^2 - AH^2} = \sqrt{8^2 - 6^2} = 2\sqrt{10}$ (cm)
- $CH = AB - AH = 8 - 6 = 2$ (cm)

a) Tính AC, BC, BH, CH:

- $AC = \sqrt{AB^2 + BC^2} = \sqrt{8^2 + BC^2}$
- $BC = \dfrac{AH}{\sin B} = \dfrac{6}{\frac{3}{4}} = 8$ (cm)
- $BH = 2\sqrt{10}$ (cm)
- $CH = 2$ (cm)

b) Tính sinC, cosC, tanC, cotC:

- $\sin C = \dfrac{AC}{AB} = \dfrac{\sqrt{8^2 + BC^2}}{8}$
- $\cos C = \dfrac{BC}{AB} = \dfrac{8}{8} = 1$
- $\tan C = \dfrac{\sin C}{\cos C} = \dfrac{\sqrt{8^2 + BC^2}}{8}$
- $\cot C = \dfrac{1}{\tan C} = \dfrac{8}{\sqrt{8^2 + BC^2}}$