Question

Tam giác ABC Có A=120 ° b=6, c=4, tìm cạnh a và góc B, C

Answer 1

a² = b² + c² - 2bc.cos120⁰

⇔ a² = 76

⇔ a = \(2\sqrt{19}\)

Theo định lí sin: \(\dfrac{a}{sinA}=\dfrac{b}{sinB}=\dfrac{c}{sinC}\)

⇔ \(\dfrac{2\sqrt{19}}{sin120}=\dfrac{6}{sinB}=\dfrac{4}{sinC}\)

⇔ \(\left\{{}\begin{matrix}sinC=\dfrac{\sqrt{57}}{19}\\sinB=\dfrac{3\sqrt{57}}{38}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\widehat{B}=36^035'\\\widehat{C}=23^025'\end{matrix}\right.\)