Question

Answer 1

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)