Giải thích các bước giải:
a.Ta có:
$\sin\widehat{HBC}=\dfrac{CH}{CB}$
$\to CH=CB\cdot\sin\widehat{HBC}$
$\to CH=6\cdot\sin60^o$
$\to CH=3\sqrt3$
Ta có:
$\hat A=180^o-\hat B-\hat C=80^o$
$\to\sin\widehat{HAC}=\dfrac{CH}{CA}$
$\to \sin80^o=\dfrac{CH}{CA}$
$\to CA=\dfrac{CH}{\sin 80^o}=\dfrac{3\sqrt3}{\sin80^o}$
b.Ta có:
$S_{ABC}=\dfrac12CA\cdot CB\cdot\sin C$
$\to S_{ABC}=\dfrac12\cdot\dfrac{3\sqrt3}{\sin80^o}\cdot 6\cdot\sin 40^o$
$\to S_{ABC}=\cdot\dfrac{9\sqrt3\cdot\sin 40^o}{\sin80^o}$
