\(\dfrac{AB}{AC}=\dfrac{3}{4}\Leftrightarrow\dfrac{AB^2}{AC^2}=\dfrac{9}{16}\);\(\Leftrightarrow\)\(\dfrac{AB^2\times CH}{AC^2\times CH}=\dfrac{BC\times BH\times CH}{BC\times CH\times CH}=\dfrac{BC\times AH^2}{BC\times CH^2}=\dfrac{AH^2}{CH^2}=\dfrac{9}{16}\)
\(\Leftrightarrow\)\(\dfrac{AH}{CH}=\dfrac{3}{4}\)\(\Leftrightarrow\)\(\dfrac{15}{CH}=\dfrac{3}{4}\)\(\Leftrightarrow\)\(HC=\left(15\times4\right)\div3\)=20
\(AH^2=HC\times HB\Rightarrow HB=AH^2\div HC=15^2\div20=11,25\)
Ta có : \(\tan C=\dfrac{AB}{AC}=\dfrac{3}{4}=>\widehat{C}\approx37^o\)
\(\widehat{B}+\widehat{C}=\widehat{A}=90^o\Rightarrow\widehat{B}=90^o-\widehat{C}=90^o-37^o=53^o\)
Xét tam giác ABH có :
\(\tan B=\dfrac{AH}{HB}=>HB=\dfrac{AH}{tanB}=\dfrac{15}{tan53^o}\approx11,3\left(cm\right)\)
Xét tam giác AHC có :
\(tanC=\dfrac{AH}{HC}\Rightarrow HC=\dfrac{AH}{tanC}=\dfrac{15}{tan37^o}\approx19,9\left(cm\right)\)
Vậy độ dài HB = 11,3 cm, độ dài HC = 19,9 cm
