An equilateral triangle a couple is inscribed in a circle with centre o .find angle boc angle Coa and angle aob
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All the angles will be same
angle AOC= angle BOC=angle AOB=120 degree
angle AOC= angle BOC=angle AOB=120 degree
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see diagram.
AF = a/2 = BF = CE = BE = AD = BD
OF = EO = DO = radius of incircle
angles AFO = OFC = OEC = OEB = ODA = ODB = 90 deg
Hence all these triangles OAD, OBD, OEB, OEC, OFC, OFA are equal.
so angles at O, angle AOF, FOC, COE, EOB, BOD, DOA are all equal.
hence, the angles at O, AOB, BOC, COA are all equal and will be equal to 360 / 3 = 120 deg.
AF = a/2 = BF = CE = BE = AD = BD
OF = EO = DO = radius of incircle
angles AFO = OFC = OEC = OEB = ODA = ODB = 90 deg
Hence all these triangles OAD, OBD, OEB, OEC, OFC, OFA are equal.
so angles at O, angle AOF, FOC, COE, EOB, BOD, DOA are all equal.
hence, the angles at O, AOB, BOC, COA are all equal and will be equal to 360 / 3 = 120 deg.
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