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 Ttangents PQ and PR are drawn at a circle with centre O from an external point P prove that <QPR = 2 <OQR

Asked by Kavya Shabu(student) , on 4/3/11

Answers

go on pg 212 of our text book... u lll get it  

Posted by Sumeet Ranka(student)on 4/3/11

 this is not the same one 

Posted by Kavya Shabu(student)on 4/3/11

EXPERT ANSWER

Hi!
Here is the answer to your question.
 
Given: A circle with centre O. PQ and PR are tangents drawn from an external point P to the circle.
To prove: ∠QPR = 2∠OQR
Proof:
∠OQP = ∠ORP = 90°     (Radius is perpendicular to the tangent at point of contact)
In quadrilateral OQPR
∠QPR + ∠OQP + ∠QOR + ∠ORP = 360°
∴∠OPR + 90° + ∠QOR + 90° = 360°
⇒ ∠OPR + ∠QOR = 180°  ... (1)
In ∆OQR
OQ = OR      (Radius of the circle)
⇒ ∠ORQ = ∠OQR      (Equal sides have equal angles opposite to them)
∠QOR + ∠OQR + ∠ORQ = 180°  (angle sum property)
⇒ ∠QOR + 2∠OQR = 180°              ... (2)
From (1) and (2), we get
∠QPR + ∠QOR = ∠QOR + 2∠OQR
⇒ ∠QPR = 2∠OQR
 
Cheers!

Posted by Lalit Mehra(MeritNation Expert)on 10/3/11

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