# 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. 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 • 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
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