Đặt $BH=x$
Khi đó $AH^2=BH.CH=x.(BC-BH)=x.(25-x)$
$⇔12^2=25x-x^2$
$⇔x^2-25x+144=0$
$⇔$\(\left[ \begin{array}{l}x=16\\x=9\end{array} \right.\) (bạn tự giải nhé)
Với $x=16$
$⇒CH=9$
khi đó $AB^2=BH.BC=16.25=400⇒AB=20(cm)$
$⇒AC=\sqrt[]{BC^2-AB^2}=\sqrt[]{25^2-20^2}=15(cm)$
Khi đó $sinC=\dfrac{AB}{BC}=\dfrac{20}{25}=\dfrac{4}{5}⇒cosB=\dfrac{4}{5}$
$cosC=\dfrac{AC}{BC}=\dfrac{15}{25}=\dfrac{3}{5}⇒sinB=\dfrac{3}{5}$
$tanC=\dfrac{AB}{AC}=\dfrac{20}{15}=\dfrac{4}{3}⇒cotB=\dfrac{4}{3}$
$cotC=\dfrac{AC}{AB}=\dfrac{15}{20}=\dfrac{3}{4}⇒tanB=\dfrac{3}{4}$
Với $BH=9⇒CH=16⇒AB=15;AC=20$
Khi đó
$sinC=\dfrac{AB}{BC}=\dfrac{15}{25}=\dfrac{3}{5}⇒cosB=\dfrac{3}{5}$
$cosC=\dfrac{AC}{BC}=\dfrac{20}{25}=\dfrac{4}{5}⇒sinB=\dfrac{4}{5}$
$tanC=\dfrac{AB}{AC}=\dfrac{15}{20}=\dfrac{3}{4}⇒cotB=\dfrac{3}{4}$
$cotC=\dfrac{AC}{AB}=\dfrac{20}{15}=\dfrac{4}{3}⇒tanB=\dfrac{4}{3}$
