As shown in the figure below,

O is the midpoint of AC.

Facing SAC⊥ facing ABC,

SO⊥AC，

∴

SO⊥ airplane ABC. If ND⊥BO is in O, then the ND⊥ plane ABC and D are the midpoint of BO, and it is not difficult to get CN=3, MN=√3 and CM=2√3.

∴

CN⊥MN，

∴

△ area of △CMN =3√3/2. In fig. 2,

△CMD area =0.5×△CMP area =0.5×(0.5×△CMB area) =√3/2.

∵

△CMD is the projective figure of △CMN on the surface ABC, let the dihedral angle be θ, and from the area projective theorem, cos θ = the area of △ cmd/the area of △ cmn = 1/3.

㈢

ND=SO/2=√2，

Let the distance from b to △CMN be h,

∵

The volume of the triangular pyramid B-CMN = the volume of the triangular pyramid N-BCM,

∴

△cmn×h area =×△BCM×nd area,

∴

(3√3/2)h=2√3×√2=4√2/3。