a) Áp dụng ĐL cos ta có:\(BC=\sqrt{AB^2+AC^2-2\cdot AB\cdot AC\cdot cosA}=\sqrt{25^2+20^2-2\cdot25\cdot20\cdot cos60^o}\approx22.91\left(cm\right)\)\(\Rightarrow\) Sb) Ta có: \(S_{ABC}=\dfrac{1}{2}\cdot AB\cdot AC\cdot sinA=\dfrac{1}{2}\cdot25\cdot20\cdot sin60^o\approx216.51\left(cm^2\right)\)\(\Rightarrow\) Đc) Ta có: Nửa chu vi \(\Delta ABC\) là : \(p=\dfrac{AB+AC+BC}{2}=\dfrac{25+20+22,91}{2}\approx33.96\)\(r=\dfrac{S}{p}=\dfrac{216,51}{33,96}\approx6.38\left(cm\right)\)\(\Rightarrow\) Đd) Ta có: \(S_{ABC}=\dfrac{1}{2}\cdot BC\cdot h_A\Rightarrow h_A=\dfrac{2S_{ABC}}{BC}=\dfrac{2\cdot216,51}{22,91}\approx18,90\) (cm)\(\Rightarrow\) SCâu trả lời: a) Áp dụng ĐL cos ta có:\(BC=\sqrt{AB^2+AC^2-2\cdot AB\cdot AC\cdot cosA}=\sqrt{25^2+20^2-2\cdot25\cdot20\cdot cos60^o}\approx22.91\left(cm\right)\)\(\Rightarrow\) Sb) Ta có: \(S_{ABC}=\dfrac{1}{2}\cdot AB\cdot AC\cdot sinA=\dfrac{1}{2}\cdot25\cdot20\cdot sin60^o\approx216.51\left(cm^2\right)\)\(\Rightarrow\) Đc) Ta có: Nửa chu vi \(\Delta ABC\) là : \(p=\dfrac{AB+AC+BC}{2}=\dfrac{25+20+22,91}{2}\approx33.96\)\(r=\dfrac{S}{p}=\dfrac{216,51}{33,96}\approx6.38\left(cm\right)\)\(\Rightarrow\) Đd) Ta có: \(S_{ABC}=\dfrac{1}{2}\cdot BC\cdot h_A\Rightarrow h_A=\dfrac{2S_{ABC}}{BC}=\dfrac{2\cdot216,51}{22,91}\approx18,90\) (cm)\(\Rightarrow\) S