Find the area of the smallest part of a disc of radius 1 0 c m , cut off by a chord A B which subtends an angle of 2 2 . 5 ∘ at the circumference.
This problem is part of the set Trigonometry .
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We know that the angle subtended by a chord on the major arc is half of the angle subtended by it at the centre. ∴ ∠ A O B = 4 5 ∘ .
Now all we have to do is find the area of the segment.
Area of Δ A O B = 2 1 a b sin θ = ( 2 1 ) ( 1 0 ) ( 1 0 ) ( sin ( 4 5 ∘ ) ) = 2 5 2
Area of sector O − A B = 3 6 0 ∘ θ × π r 2 = 3 6 0 ∘ 4 5 ∘ × π × 1 0 0 = 2 2 5 π
Hence the answer, 2 2 5 π − 2 5 2 = 4 5 0 π − 2 5 0 = 5 0 ( 4 π − 2 1 )