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find the area of segment with a 60 degree angle and 12 radius.
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\[r ^{2}(\pi \Theta \div360-\sin \Theta \div2)\]
Area of Segment = ½ × ( (θ × π/180) - sin θ) × r2 (when θ is in degrees)
Why 1/2 @maheshmeghwal9
24pi-36sqrt3?
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but it is the area of sector I think @ParthKohli
we have to find segment
's area
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