Đáp án:
Ta có:
$\hat A=180^o-\hat B-\hat C=75^o$
Hay `hatA = 180^o - 65^o - 40^o`
`⇒ hatA = 75^o`
Áp dụng định lý sin cho $Δ ABC$ ta có:
$\dfrac{BC}{\sin A}=\dfrac{CA}{\sin B}=\dfrac{AB}{\sin C}$
Hay $ \dfrac{4.2}{\sin75^o}=\dfrac{CA}{\sin65^o}=\dfrac{AB}{\sin40^o}$
Vậy
$+ CA=\dfrac{4.2\cdot\sin65^o}{\sin75^o}$
$+ AB=\dfrac{4.2\cdot\sin40^o}{\sin75^o}$
`(TM)`