Đặt $BH=x$ khi đó ta có:
$AB^2=BH.BC=x.(BH+CH)=x.(x+4)$
$⇒x.(x+4)=6^2$
$⇒x^2+4x-36=0$
$⇒(x+2)^2=40$
$⇔$\(\left[ \begin{array}{l}x+2=\sqrt[]{40}\\x+2=-\sqrt[]{40}\end{array} \right.\)
Mà $x+2>0$
$⇒x=2.\sqrt[]{10}-2$
$⇒BH=2.\sqrt[]{10}-2$
$⇒BC=2.\sqrt[]{10}-2+4=2\sqrt[]{10}+2$
$⇒AC=\sqrt[]{BC^2-AB^2}=\sqrt[]{(2\sqrt[]{10}+2)^2-6^2}=5,77(cm)$
$⇒sinB=\dfrac{AC}{BC}=\dfrac{5,77}{2.\sqrt[]10+2}$
$cosB=\dfrac{AB}{BC}=\dfrac{6}{2.\sqrt[]10+2}$
$tanB=\dfrac{\dfrac{5,77}{2.\sqrt[]10+2}}{\dfrac{6}{2.\sqrt[]10+2}}=\dfrac{5,77}{6}$
$⇒cotB=\dfrac{6}{5,77}$
$⇒tanC=\dfrac{6}{5,77}$
