Question

Cho tam giác ABC. Tính đường cao vẽ từ A và bán kính đường tròn ngoại tiếp tam giác biế 

     a. CA = 8 ; AB = 5 ; góc A bằng  

     b. BC = 21 ; CA = 17 ; AB = 8 

Answer 1

a.

\(BC=\sqrt{AB^2+AC^2-2AB.AC.cosA}=7\)

\(S=\dfrac{1}{2}AB.AC.sinA=10\sqrt{3}\)

\(\Rightarrow h_a=\dfrac{2S}{BC}=\dfrac{20\sqrt{3}}{7}\)

\(R=\dfrac{BC}{2sinA}=\dfrac{7\sqrt{3}}{3}\)

b.

\(cosA=\dfrac{AB^2+AC^2-BC^2}{2AB.AC}=-\dfrac{11}{34}\)

\(\Rightarrow sinA=\dfrac{3\sqrt{115}}{34}\)

\(S=\dfrac{1}{2}AB.AC.sinA=6\sqrt{115}\)

\(h_a=\dfrac{2S}{BC}=\dfrac{4\sqrt{115}}{7}\)

\(R=\dfrac{BC}{2sinA}=...\)