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In the given figure, AB is a chord of acircle with centre O. If POQ=2AB, prove thatangle AOB= 60°. |
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Answer» AB is a chord of circle with centre O as shown in figure.according to question, PQ passing through centre O. hence, PQ is a diameter of circle { as we know, a line segment passing through centre cut the circle at two points is known as diameter of circle } so, PQ is diameter of circle and we know, half of diameter of circle is known as radius of circle. here POQ = 2AB , means Length of AB is equal to length of radius. Let radius of circle is r then, AB = r from figure it is clear that OA and OB are the radius of circle .so, OA = OB = r now, ∆OAB, OA = OB = AB = r . all sides of ∆OAB are same so, ∆OAB is equilateral triangle .hence, all angles of ∆OAB will be 60° hence, ∠AOB = 60° |
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