In fig Op is equal to diameter of the circle. Prove that abp is a equilateral triangle..

sry abt the figure but i dnt know how to upload the figure

But its a circle with two tangents AP and BP... and joining OP, OA and OB ( o is the centre of the circle... )

AP is the tangent to the circle.

∴ OA ⊥ AP (Radius is perpendicular to the tangent at the point of contact)

⇒ ∠ OAP = 90º

In Δ OAP,


 

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let r be the radius of the circle

  AO= r

  OP=2r

 angleOAP=90'

 in right triangleOAP

  sin tita=AO/OP =r/2r =1/2

  tita=30'

therefore angle APO=30'

angleAPO =angleBPO =30'

angleAPB =30'+30' =60'

PA =PB (TANGENTS THEOREM)

ANGLE PAB= ANGLE PBA

angle PAB+ angle PBA + angle APB =180'  (ANGLE SUM PROPERTY)

angle PAB+ angle PBA +60' =180'

angle PAB+ angle PBA =120'

angle PAB= angle PBA =60'

there fore triangle APB  equilateral triangle as all anglas are 60'.

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