∴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).