Giải thích các bước giải:
a.Ta có : $\Delta ABC$ vuông tại $A$
$\to BC^2=AB^2+AC^2=225\to BC=15$
Mà $AH\perp BC\to AB\cdot AC=AH\cdot BC=2S_{ABC}$
$\to AH=\dfrac{AB\cdot AC}{BC}=\dfrac{36}{5}$
$\to BH^2=AB^2-AH^2=\dfrac{729}{25}\to BH=\dfrac{27}{5}$
b.Ta có:
$\sin\hat B=\dfrac{AC}{BC}=\dfrac45$
$\to \hat B=\arcsin\dfrac45\sim53^o7'$
c.Ta có $BD//AC\to BD\perp AB$
$\to \widehat{AHB}=\widehat{ABD}=90^o$
Mà $\widehat{BAH}=\widehat{BAD}$
$\to\Delta ABH\sim\Delta ADB(g.g)$
$\to \dfrac{AB}{AD}=\dfrac{AH}{AB}$
$\to AB^2=AH\cdot AD$
Lại có $\widehat{AHB}=\widehat{BAC}=90^o,\widehat{ABH}=\widehat{ABC}$
$\to \Delta ABH\sim\Delta CBA(g.g)$
$\to \dfrac{BA}{BC}=\dfrac{BH}{BA}$
$\to AB^2=BH\cdot BC$
$\to BH\cdot BC=AH\cdot AD$
d.Ta có: $\widehat{AHC}=\widehat{BAC}=90^o,\widehat{ACH}=\widehat{ACB}$
$\to \Delta CHA\sim\Delta CAB(g.g)$
$\to \dfrac{CH}{CA}=\dfrac{CA}{CB}$
$\to CH\cdot CB=CA^2$
$\to \dfrac{CH}{CB}=(\dfrac{CA}{CB})^2=\cos^2C$
