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Concept Detail
Maths / Circles / Angle at the Centre is Double The Angle at Circumference
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Angle at the Centre is Double The Angle at Circumference

Angle at the centre is double The Angle at circumference.

The angle subtended by an arc at the centre is twice the angle subtended at the circumference

The angle at the centre is double the angle at the circumference.

Given : Given a Circle with centre O and a Chord HK

To Prove large angle HOK=2angle HGK

Construction: Join OG, OH and OK. Extend HO to J 

Proof  Let large angle OGH=y^0 and large large angle OGK=x^0.

   Delta OGHis an isosceles. LengthsOH and OG are both radii of the same circle

angle OGH=angle OHG = y^0  [

Delta OGKis an isosceles. Lengths  OG and OK are both radii of the same circle.

angle OGK=angle OKG = x^0  [

large angle GOH=180-2y                  [ Angle sum property ]   ..............(1)

large angle GOK=180-2x                  [ Angle sum property ] ..............(2)

large angle HOK + angle GOK+angle GOH=360^0          [Angles around a point]                                                                                            ..............(3)

Putting the values from (1) and (2) in (3) we Get

large angle HOK +180-2x^0+180-2y^0=360^0

large angle HOK +360-2x^0 -2y^0=360^0

large angle HOK =360^0-360+2x^0 +2y^0

large angle HOK =2(x^0 +y^0)            ..............(4)      

large angle HGK =angle HGO+ angle OGK= x^0 +y^0   ..............(5)

From (4) and (5 ) we Get

large angle HOK =2angle HGK

Hence Proved.

Illustration:   If A, B, M are three points on a circle with centre O such that angle AOM=70^0 and  angle MOB=140^o, Find   angle AMB

 Solution:   If  A, B, M are three points on a circle with centre O such that angle AOM=70^0 and  angle MOB=140^o, 

 angle AOB = angle AOM + angle MOB = 70^0 +140^0 = 210^0

 Reflex of angle AOB=360^0- 210^0=150^0

angle AMB=frac{150^0}{2}=75^0

As angle at the centre is double the angle and the circumference.


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