Kẻ đg cao BH
+ \(sinA=\frac{BH}{AB}\Rightarrow\frac{BH}{5}=\frac{\sqrt{3}}{2}\)
\(\Rightarrow BH=\frac{5\sqrt{3}}{2}\) ( cm )
+ \(cosA=\frac{AH}{AB}\Rightarrow\frac{AH}{5}=\frac{1}{2}\) \(\Rightarrow AH=\frac{5}{2}\)
\(\Rightarrow CH=8-\frac{5}{2}=\frac{11}{2}\)
+ ΔBHC vuông tại H
\(\Rightarrow BC=\sqrt{BH^2+CH^2}=\sqrt{\frac{75}{4}+\frac{121}{4}}=7\) (cm)
Kẻ đường cao BH
Xét \(\Delta ABH\) vuông tại H, \(\widehat{A}=60^0\)
\(\Rightarrow BH=AB.\sin A=5.\sin60^0=\frac{5\sqrt{3}}{2}\left(cm\right)\)
\(\Rightarrow AH=AB.\cos A=5.\cos60^0=\frac{5}{2}\left(cm\right)\)
\(\Rightarrow HC=AC-AH=8-\frac{5}{2}=\frac{11}{2}\left(cm\right)\)
Xét \(\Delta BHC\) vuông tại H
Theo đly Py-ta-go:
\(BC^2=BH^2+HC^2\Rightarrow BC=\sqrt{BH^2+HC^2}=\sqrt{49}=7\left(cm\right)\)
