A chord of a circle of radius 15cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments of the circle. (Use π=3.14 and √3=1.73)

In the mentioned figure, O is the centre of the circle, AB is a chord, AXB is a minor arc, OA=OB=radius=15cm. Arc AXB subtends an angle 60° at O.
Area of sector = θ/360 × π × r² where, θ=central angle, r=radius
Area of sector AOB = 60/360 × π × r² = 60/360 × 3.14 × (15)² = 117.75cm²
OC will bisect ∠AOB, you can get this, as △AOC ≅ △BOC are congruent by RHS congruence Rule.
By trigonometry, we have
AC = 15sin30 [sinθ=P/H]
OC = 15cos30 [cosθ=B/H]
And, AB = 2AC ∴AB = 2 × 15sin30 = 15cm
∴OC = 15cos30 = 15√3/2 = 15 × 1.73/2 = 12.975cm
∴Area of △AOB = 0.5 × 15 × 12.975 = 97.3125cm² [ Area of △ = 1/2 × Base × height]
Area of the minor segment (Area of Shaded region) = Area of sector AOB - Area of △AOB
∴Area of minor segment (Area of Shaded region) = 117.75 - 97.3125 = 20.4375cm²
Area of major segment = Area of circle - Area of the minor segment = (3.14 × 15 × 15) - 20.4375 = 686.0625cm²