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Two tangents tq and tp are …

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Two tangents tq and tp are drawn to a circle with centre o from an external point p prove that anglePTQ =2angle OPQ

Posted by Dev Raj 6 years, 9 months ago

CBSE > Class 10 > Mathematics

4 answers

Gaurav Seth 6 years, 9 months ago

We know that, the lengths of tangents drawn from an external point to a circle are equal.

∴ TP = TQ

In ΔTPQ,

TP = TQ

⇒ ∠TQP = ∠TPQ ...(1) (In a triangle, equal sides have equal angles opposite to them)

∠TQP + ∠TPQ + ∠PTQ = 180º (Angle sum property)

∴ 2 ∠TPQ + ∠PTQ = 180º (Using(1))

⇒ ∠PTQ = 180º – 2 ∠TPQ ...(1)

We know that, a tangent to a circle is perpendicular to the radius through the point of contact.

OP ⊥ PT,

∴ ∠OPT = 90º

⇒ ∠OPQ + ∠TPQ = 90º

⇒ ∠OPQ = 90º – ∠TPQ

⇒ 2∠OPQ = 2(90º – ∠TPQ) = 180º – 2 ∠TPQ ...(2)

From (1) and (2), we get

∠PTQ = 2∠OPQ

Vikram Singh 6 years, 9 months ago

XD sorry for the wrong answer...

Parneet Kaur 6 years, 9 months ago

Tangents-pq and pr from p Proof-join oq and or Oq=or(radii) Angle oqr=ANGLE orq(angles Opp to equal sides) So,angle pqr =90°-Angle oqr Ang prq=90°- orq=90°-oqr So,pqr+prq=90°-oqr+90°-oqr Pqr+prq=180°-2 angle oqr 180°- qpr=180 ° - 2angle oqr(a.s.p of triangle) Angle qpr=2angle oqr

Vikram Singh 6 years, 9 months ago

Check question again its insufficient to prove..angle is missing

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