如图所示,在三棱锥V-ABC中,VC垂直底面ABC,AC垂直BC,D是AB的中点,且AC=BC=a,角VDC=D(0

易知AC,BC,VC两两垂直
建立直角坐标系如图,
因为AC = BC,D为AB中点,所以CD垂直AB,CD = a/sqrt(2),而VC垂直面CAB,故角VCD = 90, 设面VAB的法向量t为(x,y,z),t垂直于向量VA,VB,而角VDC = D,所以VC = a*tan(D)/sqrt(2), 故VA = (a,0,-a*tan(D)/sqrt(2)),VB = (0,a,-a*tan(D)/sqrt(2)),
由于t*VA = t*VB = 0,可知,t = (tan(D)/sqrt(2),tan(D)/sqrt(2),1),而CB = (0,a,0),故CB与面VAB的夹角α满足:
α = 90 -
故：
sin(α) = cos() = t*CB/||t||/||CB|| = tan(D)/sqrt(2)/sqrt(tanD*tanD+1) = sin(D)/sqrt(2)
根据sin(D)在0-90上的单调性,0