Verified answer

The smaller circle meets line segment AB at an unlabeled point. Let me call it E. The shaded region can be cut an common chord CD into two circle segments: segment CBD on the large circle, and segment CED on the small circle.

area(segment CBD)

= area(sector CBD) - area(∆ACD)

= 1/2(α)(2r)² - 1/2(2r)²sin(α)

= 2r²[α - sin(α)]

By the same process,

area(segment CED) = 1/2r²[θ - sin(θ)]

area(shaded region) = 2r²[α - sin(α)] + 1/2r²[θ - sin(θ)]

Draw BC, and let M be its midpoint. Line AB is the bisector of ∠CAD, and line AM is the bisector of ∠BAC.

∠BAM = α/4

∠BMA = 90°

BC = r

BM = r/2

AB = 2r

sin(∠BAM) = BM/AB

sin(α/4) = (r/2)/(2r)

sin(α/4) = 1/4

α/4 = arcsin(1/4)

α = 4arcsin(1/4)

For the last part, it looks like you need to express θ in terms of α. For that I would use isosceles triangle ABC.

∠BAC = α/2

∠ABC = ∠ACB = θ/2

You will then have α, θ, and the area known. That leave r as the only remaining unknown in that equation in part (a).