Question

[Trần Nhật Quang]

Cho tam giac ABC co A bang 60 do.AB= 5cm,AC=8cm .Tinh BC

Answer 1

[bach nhac lam]

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)

Answer 2

[Hoàng Tử Hà]

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:

\(BC^2=BH^2+HC^2\Rightarrow BC=\sqrt{BH^2+HC^2}=\sqrt{49}=7\left(cm\right)\)