Question

Cho tam giác ABC . Hãy tính cạnh BC biết

a, AB = 1cm , AC= 2cm , góc BAC = 120 độ

b, AB = 1dm , AC = 5cm , góc BAC = 60 độ

c, AB= 2cm ,AC= \(\sqrt{3}\)cm , BAC = 60 độ

Answer 1

a: \(\cos BAC=\dfrac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}=\dfrac{5-BC^2}{2\cdot1\cdot2}=\dfrac{5-BC^2}{4}\)

\(\Leftrightarrow\dfrac{5-BC^2}{4}=\dfrac{-1}{2}\)

\(\Leftrightarrow5-BC^2=-2\)

\(\Leftrightarrow BC=\sqrt{7}\left(cm\right)\)

b: \(\cos BAC=\dfrac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}=\dfrac{125-BC^2}{100}\)

\(\Leftrightarrow125-BC^2=50\)

hay \(BC=5\sqrt{3}\left(cm\right)\)

c: \(\cos BAC=\dfrac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}=\dfrac{7-BC^2}{8\sqrt{3}}\)

\(\Leftrightarrow7-BC^2=4\sqrt{3}\)

hay \(BC=2-\sqrt{3}\left(cm\right)\)