Question

Cho tam giác ABC biết \(c=35cm,\widehat{A}=40^0;\widehat{C}=120^0\). Tính \(a,b,\widehat{B}\) ?

Answer 1

Answer 2

\(\widehat{B}=180^o-\left(40^o+120^o\right)=20^o\).

\(AH=AB.sinB=35.sin20^o\cong12cm.\)
\(\widehat{HCA}=180^o-120^o=60^o\).
\(AH=AC.sin60^o\Rightarrow AC=\dfrac{AH}{sin60}=\dfrac{12}{\dfrac{\sqrt{3}}{2}}=8\sqrt{3}\).
Áp dụng định lý Cô-sin:
\(BC=\sqrt{AB^2+AC^2-2.AB.AC.sinA}\)\(=\sqrt{35^2+\left(8\sqrt{3}\right)^2-2.35.8\sqrt{3}.cos40^o}\cong26cm\).
Vậy \(a=26cm;b=8\sqrt{3}cm,\)\(\widehat{B}=20^o\).