Question

[Đỗ Minh Ngọc]

Tam giác $ABC$ có $\hat{A}={60}^0,\hat{B}={45}^0,b=4$. Tính hai cạnh $a$ và $c$.
 

Answer 1

[Đoàn Đức Hà]

\(\widehat{C}=180^o-\widehat{A}-\widehat{B}=180^o-60^o-45^o=75^o\)

Theo định lí hàm \(sin\)trong tam giác: 

\(\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}\)

\(\Rightarrow\hept{\begin{cases}a=\frac{bsinA}{sinB}=\frac{4.sin60^o}{sin45^o}=2\sqrt{6}\\c=\frac{bsinC}{sinB}=\frac{4sin60^o}{sin75^o}=-2\sqrt{6}+6\sqrt{2}\end{cases}}\)