#### Question

Tangents AP and AQ are drawn to a circle, with centre O, from an exterior point A. prove that: PAQ = 2`∠`OPQ

#### Solution

In quadrilateral OPAQ,

`∠`OPA = `∠` OQA = 90º

(∵ OP ⊥ PA and OQ ⊥ QA)

∴ `∠`POQ + `∠`PAQ +90° +90°=360°

⇒ `∠`POQ +`∠`PAQ =360° -180°=180° ........................(i)

In triangle OPQ,

OP = OQ (Radii of the same circle)

∴ OPQ = `∠`OQP

But

`∠`POQ + `∠` OPQ + `∠` OQP =180°

⇒ `∠`POQ + `∠`OPQ +`∠`OPQ = 180°

⇒ `∠`POQ + 2 `∠`OPQ = 180° ...............................(ii)

From (i) and (ii)

`∠`POQ + `∠`PAQ = `∠`POQ + 2 OPQ

⇒ `∠` PAQ = 2 `∠` OPQ

Is there an error in this question or solution?

Solution Tangents Ap and Aq Are Drawn to a Circle, with Centre O, from an Exterior Point A. Prove That: Paq = 2`∠`Opq Concept: Tangent Properties - If a Line Touches a Circle and from the Point of Contact, a Chord is Drawn, the Angles Between the Tangent and the Chord Are Respectively Equal to the Angles in the Corresponding Alternate Segments.