If from a external point P of a circle with centre O, two tangents PQ and PRare drawn such that angle of QPR=120. prove tht 2PQ=PO.

Here is the answer to your query.



Given : BC and BD are tangents with ∠DBC = 120°


 (tangent from an external point are equally inclined to the segment joining the centre to that point)


Now in ΔBCO

⇒ OB = 2BC = BC + BC

⇒ OB = BC + BD ( BC = BD as tangents drawn from an external point to a circle are equal)

  • 30
@Mohil but how will we prove that PQ+PR=PO????
  • -3

In  ▲OPQ, we have
∠PQO=90⁰ (the tangent at any point is perpendicular to the radius through the point of contact) and ∠QPO=1/2 X ∠QPR= 1/2 X 120⁰= 60⁰ (the two tangents drawn from an external point are equally inclined to the line segment joining the centre to that point and so ∠QPO=∠RPO) In right ▲OPQ, we have cos(∠QPO)=PQ/PO COS 60⁰ = PQ/PO 1/2=PQ/PO 2PQ=PO HENCE PROVED HOPE THIS HELPS....

  • 1
How can you proove
  • -5
Please find this answer

  • 39
good morning students,
students are confused mainly becoz of using cos@ and sin@.here  we need BC and BO so we can apply cos@ easily. this is the main reason behind the problem b/h=BC/OB then u will get the ans easily.that means we have done the problem according to the question.
                                                                                  cheers ,
                                                                           hope this helps
  • -8
In your question there is hell of spelling mistake.First learn english then mathematics okay.
  • -3
McDonald's has provided the best way to do the question. Appreciate thou.
  • 0
What are you looking for?