Giải thích các bước giải:
Gọi $AD\perp BC\to \tan\hat B=\dfrac{AD}{BD},\tan \hat C=\dfrac{AD}{DC}$
$\to BD=\dfrac{AD}{\tan 40^o}, CD=\dfrac{AD}{\tan 55^o}$
$\to BC=BD+CD=\dfrac{AD}{\tan 40^o}+\dfrac{AD}{\tan 55^o}$
$\to AD(\dfrac{1}{\tan 40^o}+\dfrac{1}{\tan 55^o})=40$
$\to AD=40.\dfrac{\tan \left(40^o\right)\tan \left(55^o\right)}{\tan \left(55^o\right)+\tan \left(40^o\right)}$
$\to S_{ABC}=\dfrac{1}{2}AD.BC=800.\dfrac{\tan \left(40^o\right)\tan \left(55^o\right)}{\tan \left(55^o\right)+\tan \left(40^o\right)}$
