Đáp án: $ S_{ABC}=\dfrac{800}{\dfrac{1}{\tan40^o}+\dfrac{1}{\tan55^o}}$
Giải thích các bước giải:
Kẻ $AH\perp BC=H$
Ta có:
$\tan\widehat{ABH}=\dfrac{AH}{BH}$
$\to \tan40^o=\dfrac{AH}{BH}$
$\to BH=\dfrac{AH}{\tan40^o}$
Lại có:
$\tan\widehat{ACH}=\dfrac{AH}{CH}$
$\to\tan55^o=\dfrac{AH}{CH}$
$\to CH=\dfrac{AH}{\tan55^o}$
Mà $BC=40$
$\to BH+HC=40$
$\to \dfrac{AH}{\tan40^o}+\dfrac{AH}{\tan55^o}=40$
$\to AH(\dfrac{1}{\tan40^o}+\dfrac{1}{\tan55^o})=40$
$\to AH=\dfrac{40}{\dfrac{1}{\tan40^o}+\dfrac{1}{\tan55^o}}$
$\to S_{ABC}=\dfrac12\cdot AH\cdot BC$
$\to S_{ABC}=\dfrac12\cdot \dfrac{40}{\dfrac{1}{\tan40^o}+\dfrac{1}{\tan55^o}}\cdot 40$
$\to S_{ABC}=\dfrac{800}{\dfrac{1}{\tan40^o}+\dfrac{1}{\tan55^o}}$
