圆（x-3）^2+(y+4)^2=4和直线y=kx相交于点P,Q
y=kx
（x-3）^2+(y+4)^2=4
（x-3）^2+(KX+4)^2=4
(1+k^2)x^2+(8k-6)x+21=0
x(P)*x(Q)=21/(1+k^2)
y(P)=k*x(P),y(Q)=k*x(Q)
OP^2=[x(P)]^2+[y(P)]^2=(1+k^2)*[x(P)]^2
OQ^2=[x(Q)]^2+[y(Q)]^2=(1+k^2)*[x(Q)]^2
(OP*OQ)^2=[x(P)*x(Q)]^2*(1+k^2)^2=[21/(1+k^2)]^2*(1+k^2)^2=21^2
OP*OQ=21如图,D是A在面内的射影,
E是C在面内的射影过A作AF⊥BC于F,
则面ADEC与面α垂直,故AC在面内的射影即DE,
直线AC与面α的夹角即AC与DE所成的锐角由作图知,∠CAF的大小即即线面角的大小,
由已知及作图,AB=3,BC=42,∠ABD=30°,∠CBE=45°
∴AD=32,CE=4,
由作图知CF=52,又AC=5,
在直角三角形AFC中,sin∠CAF=525=12,
∴∠CAF=30°,即AC与面α所成的角是30°．