a) Ta có:
$\tan\widehat{ABC} = \dfrac{AC}{AB} = \dfrac{AC}{\dfrac{1}{3}AC} = 3$
$\Rightarrow \widehat{ABC} = \arctan3 \approx 71^o34'$
$\Rightarrow \widehat{BCA} = 90^o - \widehat{ABC} = 90^o - 71^o34' = 18^o26'$
b) Xét $ΔABH$ và $ΔCAH$ có:
$\widehat{AHB} = \widehat{AHC} = 90^o$
$\widehat{HAB} = \widehat{HCA}$ (cùng phụ $\widehat{HAC}$)
Do đó $ΔABH \sim ΔCAH \, (g.g)$
$\Rightarrow \dfrac{AB}{AC} = \dfrac{AH}{CH}$
$\Rightarrow \dfrac{AB^2}{AC^2} = \dfrac{AH^2}{CH^2} = \dfrac{BH.CH}{CH^2} = \dfrac{BH}{CH}$
$\Rightarrow \dfrac{BH}{CH} = \dfrac{AB^2}{AC^2} = \dfrac{\dfrac{1}{9}AC^2}{AC^2} = \dfrac{1}{9}$
c) Ta có: $ΔABH\sim ΔCAH$ (câu b)
$\Rightarrow \dfrac{S_{ABH}}{S_{CAH}} = \dfrac{AB^2}{AC^2} = \dfrac{1}{9}$
$\Rightarrow 9S_{ABH} = S_{CAH}$
mà $S_{ABH} + S_{CAH} = S_{ABC}$
nên $10S_{ABH} = S_{ABC} = 15 \, cm^2$
$\Rightarrow S_{ABH} = \dfrac{15}{10} = \dfrac{3}{2} \, cm^2$
