Lời giải:
Ta có:
$BH=\frac{AH}{\tan B}=\frac{AH}{\tan 60^0}$
$CH=\frac{AH}{\tan C}=\frac{AH}{\tan 40^0}$
$\Rightarrow BH+CH=\frac{AH}{\tan 60^0}+\frac{AH}{\tan 40^0}$
$\Leftrightarrow 12=AH(\frac{1}{\tan 60^0}+\frac{1}{\tan 40^0})$
$\Rightarrow AH=6,78$ (cm)
$AC=\frac{AH}{\sin C}=\frac{AH}{\sin 40^0}=10,55$ (cm)
b.$S_{ABC}=\frac{AH.BC}{2}=\frac{6,78.12}{2}=40,68$ (cm vuông)
Đúng 0
Bình luận (0)
