Ab is the chord of a circle with centre o,if angle AOB =60°.Prove that Ab is half of diameter
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Hi Gitanjali,
Given
AB is the chord.
<AOB = 60
now in Triangle AOB,
<OAB = <OBA .....since OA and OB are of same length = radius
so
<AOB + <OAB + <OBA = 180
=> <OAB + <OBA = 180-60
=> 2<OAB = 120
=> <OAB = 60 = <OBA
all angles are 60 so AOB s an equilateral triangle
hence all sidesare are equal
so
AB = OB = OA = radius = Half of diameter
Hence proved
Given
AB is the chord.
<AOB = 60
now in Triangle AOB,
<OAB = <OBA .....since OA and OB are of same length = radius
so
<AOB + <OAB + <OBA = 180
=> <OAB + <OBA = 180-60
=> 2<OAB = 120
=> <OAB = 60 = <OBA
all angles are 60 so AOB s an equilateral triangle
hence all sidesare are equal
so
AB = OB = OA = radius = Half of diameter
Hence proved
jish4you:
hope u liked my solution
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