В треугольнике ABC угол C равен 90, CH-высота ,AC=10, tgA=корень из 21/2.найдите AH

[latex]tgA= \frac{CB}{AC}[/latex] [latex]\frac{ \sqrt{21} }{2}=\frac{CB}{10}[/latex] [latex]CB=5 \sqrt{21}[/latex] [latex]AB= \sqrt{AC^{2}+CB^{2}}= \sqrt{100+25*21}= \sqrt{625}=25[/latex] [latex]CH= \frac{AC*CB}{AB}= \frac{10*5 \sqrt{21}}{25}=2\sqrt{21}[/latex] [latex]AH= \sqrt{AC^{2}- CH^{2}}= \sqrt{100-4*21}=4 [/latex]