Give step by step explanation for the following question?

Method 1:

Refer to the figure below for Method 1.
Join OA, OB and OC. They are radii of the circle.

In ΔAOC:
AC² = OA² + OC² - 2 × OA × OC × cos(∠AOC)   (cosine law)
(R√3)² = R² + R² - 2 × R × R × cos(∠AOC)
3 = 2 - 2 cos(∠AOC)
cos(∠AOC) = -1/2
∠AOC = 120°

In ΔAOB:
AB² = OA² + OB² - 2 × OA × OB × cos(∠AOB)   (cosine law)
(R√2)² = R² + R² - 2 × R × R × cos(∠AOB)
2 = 2 - 2 cos(∠AOB)
cos(∠AOB) = 0
∠AOB = 90°

Area of minor segment AC
= (Area of sector AOC) - (Area of ΔAOC)
= π × R² × (120°/360°) - (1/2) × OA × OC × sin(∠AOC)
= (πR²/3) - (R²(√3)/4)

Area of minor segment AB
= (Area of sector AOB) - (Area of ΔAOB)
= π × R² × (90°/360°) - (1/2) × R² × sin(90°)
= (πR²/4) - (R²/2)

Area of the shaded area
= (Area of minor segment AC) - (Area of minor segment AB)
= (πR²/3 ) - [R²(√3)/4] - [(πR²/4) - (R²/2)]
= (R²/12)(π + 6 - 3√3)  (square units)
≈ 0.3288R²  (square units)

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Method 2:

Refer to the figure below for Method 2.
Join OA, OB and OC. They are radii of the circle.
Draw the altitude OH of for ΔOAB, and the altitude OK for ΔOAC.

In ΔAOK:
OA² = OK² + AK²  (Pythagorean theorem)
R² = OK² + [R(√3)/2)]²
OK = √[R² - (3R²/4)] = R/2

In ΔAOK:
sin(∠AOK) = AK/OA = [R(√3)/2)]/R = (√3)/2
∠AOK = 60°
∠AOC = 60° × 2 = 120°

In ΔAOH:
OA² = OH² + AH²  (Pythagorean theorem)
R² = OH² + [R(√2)/2)]²
OH = √[R² - (2R²/4)] = R/√2 = R(√2)/2

In ΔAOH:
sin(∠AOH) = AH/OA = [R(√2)/2)]/R = (√2)/2
∠AOH = 45°
∠AOB = 45° × 2 = 90°

Area of minor segment AC
= (Area of sector AOC) - (Area of ΔAOC)
= π × R² × (120°/360°) - (1/2) × (R/2) × (R√3)
= (πR²/3) - (R²(√3)/4)

Area of minor segment AB
= (Area of sector AOB) - (Area of ΔAOB)
= π × R² × (90°/360°) - (1/2) × [R(√2)/2] × R(√2)
= (πR²/4) - (R²/2)

Area of the shaded area
= (Area of minor segment AC) - (Area of minor segment AB)
= (πR²/3 ) - [R²(√3)/4] - [(πR²/4) - (R²/2)]
= (R²/12)(π + 6 - 3√3)  (square units)
≈ 0.3288R²  (square units)

 In the given figure, the radius of the circle is R, 
 and the lengths of the chords AC and AB 
 are R√3 and R√2 respectively.  
 ∠ AOC = 120° and ∠ AOB = 90°
 What is the area of the shaded region?
 Area of the shaded area
 = (Area of minor segment AC) - (Area of minor segment AB)
 = (πR²/3 ) - [R²(√3)/4] - [(πR²/4) - (R²/2)]
 = (π R²)/12 - (√3) R²)/4 + R²/2
 = (1/12) (π - 3√3 + 6) (R²)
 = 0.32878668590693005 (R²)
 

∠AOC = 2sin⁻¹[√(3)/2] = 2π/3

∠AOB = 2sin⁻¹[√(2)/2] = π/2

area(segment AC)
= (R²/2)(2π/3) - (R²/2)sin(2π/3)
= (R²/2)[2π/3 - √(3)/2]

area(segment AB)
= (R²/2)(π/2) - (R²/2)sin(π/2)
= (R²/2)(π/2 - 1)

area(shaded region)
= area(segment AC) - area(segment AB)
= (R²/2)[2π/3 - √(3)/2] - (R²/2)(π/2 - 1)
= (R²/2)[π/6 + 1 - √(3)/2]
= (R²/12)[π + 6 - 3√(3)]