BH=3CH nên CH=1/3BH
\(AH^2=HB\cdot HC\)
\(\Leftrightarrow HB^2\cdot\dfrac{1}{3}=AH^2\)
\(\Leftrightarrow HB^2=3\cdot AH^2\)
\(\Leftrightarrow HB=\sqrt{3}\cdot AH\)
\(\Leftrightarrow\dfrac{AH}{HB}=\dfrac{1}{\sqrt{3}}\)
\(\Leftrightarrow tanB=\dfrac{1}{\sqrt{3}}\)
=>góc B=30 độ