设$H$为垂心,且$3\overrightarrow{HA}+4\overrightarrow {HB}+5\overrightarrow {HC}=\overrightarrow 0$,则$\cos\angle AHB=$____

分析:$\tan A\overrightarrow {HA}+\tan B\overrightarrow {HB}+\tan C\overrightarrow {HC}=\overrightarrow 0$,
故$tan A:tan B:tan C=3:4:5$ 又$\tan A\tan B\tan C=\tan A\tan B\tan C $
故$(\tan A,\tan B,\tan C)=(\dfrac{3}{\sqrt{5}},\dfrac{4}{\sqrt{5}},\sqrt{5})$, 
从而$\cos\angle{AHB}=-\cos C=-\dfrac{\sqrt{6}}{6}$

练习:$\Delta ABC $中$AB=4,AC=3,BC=2$,点$H$为三角形的垂心,
若$\overrightarrow {AH}=x\overrightarrow {AB}+y\overrightarrow {AC}$则$\dfrac{y}{x}=$_____
答案:$-\dfrac{11}{3}$
另外不常用的一个外心的结论$\sin 2A \overrightarrow {OA}+\sin2B\overrightarrow {OB}+\sin2C\overrightarrow {OC}=\overrightarrow 0$