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A chord of a circle subtends an angle θ at the centre of circle. The area of the minor segment cut off by the chord is one eighth of the area of circle. Prove that 8 sin θ/2 . cos θ/2 +π = πθ/45

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Let radius of circle = r

Area of circle = πr2

AB is a chord, OA, OB are joined drop OM ⊥ AB. This OM bisects AB as well as ∠AOB.

Area of segment cut off by AB = (area of sector) – (area of triangles)

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