as level p1 maths 👍👍👍👍?

∠OBC = ∠OBA = π/6 ... (symmetry on line OB)

BC = BA = 3/tan(π/6) = 3√(3) cm

∠COA = π - ∠CBA = 2π/3 ... (opposite angles of cyclic quadrilateral)

arc AC = 3∠COA = 2π cm

(i) perimeter

= BC + BA + arc AC

= [6√(3) + 2π] cm

(ii) shaded area

= area(OABC) - area(sector OAC)

= (OA)(AB) - (OA)(arc AC)/2

= [9√(3) - 3π] cm²

Arc AC = rθ = 3 x 2π/3 = 2π (∵∠AOC = π – π/3 = 2π/3)

AB = BC = 3tan(π/3) = 3√3

(i) P = 3√3 + 3√3 + 2π = 6√3 + 2π.

(ii) Area ABCD = 2 x area ∆AOB = 2 x (1/2) x 3 x 3√3 = 9√3

Area sector AOC = (1/2)r²θ = (9/2) x 2π/3 = 3π

Area shaded region = 9√3 - 3π.

If you draw OB then it should be clear that angle CBO is 30°, and thus from the triangle BOC, the angle BOC is 60° and thus COA is 120°.

You now have all the angles of BOC and the length of CO (3cm).

That is all the info you need