Đáp án:
$HB=\dfrac{{75}}{7};HC=21$
Giải thích các bước giải:
Ta có:
$\begin{array}{l}
AC = x \Rightarrow AB = \dfrac{5}{7}x\\
\Delta ABC;\widehat {BAC} = {90^0};AH \bot BC = H\\
\Rightarrow \dfrac{1}{{A{H^2}}} = \dfrac{1}{{A{B^2}}} + \dfrac{1}{{A{C^2}}}\\
\Leftrightarrow \dfrac{1}{{{{15}^2}}} = \dfrac{1}{{{{\left( {\dfrac{5}{7}x} \right)}^2}}} + \dfrac{1}{{{x^2}}} \Leftrightarrow x = \sqrt {666} \\
\Rightarrow \left\{ \begin{array}{l}
HB = \sqrt {A{B^2} - A{H^2}} = \sqrt {{{\left( {\dfrac{5}{7}\sqrt {666} } \right)}^2} - {{15}^2}} = \dfrac{{75}}{7}\\
HC = \sqrt {A{C^2} - A{H^2}} = \sqrt {{{\left( {\sqrt {666} } \right)}^2} - {{15}^2}} = 21
\end{array} \right.
\end{array}$
Vậy $HB=\dfrac{{75}}{7};HC=21$
