a) Có \(BC=BH+CH=25\)
Áp dụng hệ thức trong tam giác vuông có:
\(AH^2=BH.CH=144\Rightarrow AH=12\)
\(AB^2=BH.BC=225\Rightarrow AB=15\)
\(AC^2=CH.BC=400\Rightarrow AC=20\)
\(sinB=\dfrac{AC}{BC}=\dfrac{20}{25}\Rightarrow\widehat{B}\approx53^0\)
\(sinC=\dfrac{AB}{BC}=\dfrac{15}{25}\Rightarrow\widehat{C}\approx37^0\)
b) Có \(\dfrac{sinB}{sinC}=\dfrac{AC}{BC}:\dfrac{AB}{BC}=\dfrac{AC}{AB}\)