MN=AB-2-2=22

2. Make DF⊥AB at the intersection D, because ∠ DAB = ∠ B = 30, ∴AD=DB, AF=FB, AD=26√3/3.

3. Connect OE, and extend DF to O. In the right triangle OED, ∠ OED = 90, ∠ ADF = 60, so the radius is solved by an equation group composed of two right triangles.