C2 question please help :(?

Question

A circle has 3 points, A, B, C, around the perimeter of the circle with centre O and radius 4cm, such that AB is a diameter of the circle and AOC=θ radians. Given that the area of the sector BOC=5cm² more than the area of the sector AOC.

(a) show that θ=(8π-5)/16

(b) calculate the difference between the arc length BC and the arc length AC.

Answer 1

Angle BOC is π-θ. The area of a sector with angle φ is (1/2) r^2 φ, so the area of sector AOC is 8θ and the area of sector BOC is 8(π-θ). So we have
8(π-θ) = 8θ + 5
=> 16θ = 8π - 5
=> θ = (8π-5)/16.

Angle BOC is π-θ = (16π-8π+5)/16 = (8π+5)/16. The arc length of a sector with angle φ is rφ, so the difference in arc length is
4(8π+5)/16 - 4(8π-5)/16
= 2(4)(5)/16
= 5/2 cm.

Answer 2

Area of a sector is computed as

A = 1/2 r^2 θ

the area of sector AOB which is a half circle, wherein θ = π is

A(AOB) = 1/2 (4^2) π
A(AOB) = 8 π

from the given below, it can be observed when drawn that

A(BOC) + A(AOC) = A (AOB)

given that A(BOC) = 5 + A(AOC), we have

5 + A(AOC) + A(AOC) = 8π

5 + 2A(AOC) = 8π

therefore A(AOC) = (8π - 5)/2,

further,

A(AOC) = 1/2 (4^2) θ = (8π - 5)/2

thus,
θ = (8π - 5)/16


b) 2.5 cm

S = r θ

S(AC) = 4 * (8π - 5)/16 = (8π - 5)/4

but S(ACB) = S(AC) + S(BC), and S(ACB) is 4π thus,

S(BC) = 4π - (8π - 5)/4 = (8π + 5)/4

therefore

S(BC) - S(AC) = (8π + 5)/4 - (8π - 5)/4 = 10/4 = 2.5