Question

Pq and pr are tangents to the circle centre o. ps and so are equal. prove that pqr is an equilateral

Answer 1

heya it's Aastha!!!!!!!!!!!


PA and PB are the tangents to the circle.

∴ OA ⊥ PA

⇒ ∠OAP = 90°

In ΔOPA,

 sin ∠OPA = OA OP  =  r 2r  

⇒ sin ∠OPA = 1 2 =  sin  30 ⁰


⇒ ∠OPA = 30°

Similarly, it can be proved that ∠OPB = 30°.

Now, ∠APB = ∠OPA + ∠OPB = 30° + 30° = 60°

In ΔPAB,

PA = PB        

⇒∠PAB = ∠PBA ............(1)   

∠PAB + ∠PBA + ∠APB = 180°    

⇒∠PAB + ∠PAB = 180° – 60° = 120°  [Using (1)]

⇒2∠PAB = 120°

⇒∠PAB = 60°    .............(2)

From (1) and (2)

∠PAB = ∠PBA = ∠APB = 60°

∴ ΔPAB is an equilateral triangle.

hope it help you!!!!!!!!

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