Let point a be AH⊥A 1B of point h, and connect CH.

∫ plane A 1BC⊥ plane A 1ABB 1, plane A 1BC∩ plane a1abb1= a1b.

∴AH⊥ Aircraft A 1BC

So ∠ACH is the angle formed by the straight line AC and the plane A 1BC, so sin θ = ah/AC.

∫ plane ABC⊥ plane A 1ABB 1, plane ABC∩ plane A 1ABB 1 = AB, BC⊥AB.

∴BC⊥ plane a 1ab 1

∴BC⊥A 1B

So ∠ABH is the plane angle of dihedral angle a 1-BC-a, so sin φ = ah/ab.

Because AC > ab

∴sinθ