Solution: Choose A, as shown in figure 1. In △BCD, we can get ∠CBD=π-α-β, BCsin ∠ BDC = CDsin ∠ CBD from sine theorem.

∴BC=ssinβsin(α+β)

In the right angle △ABC, ab = bctan ∠ ACB = Stan θ sin β sin (α+β);

As shown in fig. 2, option b,

At △AEF, ∠EAF=β-α, EFsin(β? α)=AFsinα

∴AF=ssinαsin(β？ α)

In the right angle △ABF, AB=AFsinβ=ssinαsinβsin(β? α).