Xét tam giác ABC vuông tại A có AH là đường cao:
+) \(\tan C=\dfrac{AB}{AC}\) (TSLG)
\(\Rightarrow\tan C=\dfrac{3}{4}\Rightarrow\widehat{C}\approx37^0\)
\(\Rightarrow\widehat{B}=90^0-\widehat{C}\approx90^0-37^0\approx53^0\)
+) \(\sin C=\dfrac{AB}{BC}\) (TSLG)
\(\Rightarrow\sin37^0=\dfrac{AB}{20}\Rightarrow AB\approx12\) (cm)
+) \(AB^2+AC^2=BC^2\) (ĐL Pytago)
\(\Rightarrow AC=\sqrt{BC^2-AB^2}\approx\sqrt{20^2-12^2}\approx16\) (cm)
+) \(AB^2=BH.BC\) (HTL)
\(\Rightarrow BH=\dfrac{AB^2}{BC}\approx\dfrac{12^2}{20}\approx7,2\) (cm)
+) \(BH+CH=BC\)
\(\Rightarrow CH=BC-BH\approx20-7,2\approx12,8\) (cm)
Vậy \(HB\approx7,2cm;HC\approx12,8cm\)
