若方程（m+5）x^2-（2m+5）x+12=0的两根是直角三角形两锐角的正弦值,则m=?

x1=sina x2=sinb=sin(90-a)=cosa
x1^2+x2^2=sina^2+cosa^2=1
x1+x2=(2m-5)/(m+5) x1x2=12/(m+5)
x1^2+x2^2=(x1+x2)^2-2x1x2=(2m-5)^2/(m+5)^2-24/(m+5)=1
(2m-5)^2-24(m+5)=(m+5)^2
4m^2-20m+25-24m-120=m^2+10m+25
3m^2-54m-120=0
(3m-60)(m+2)=0
m>0
m=60/3=20