#### Question

In the following figure, PQ and PR are tangents to the circle, with centre O. If `∠`QPR = 60°, calculate:

(i) ∠QOR (ii) `∠`OQR (iii) `∠`QSR

#### Solution

Join QR.

i) In quadrilateral ORPQ,

OQ ⊥ OP,OR ⊥ RP

∴ `∠`OQP = 90° ,`∠`ORP = 90° ,`∠`QPR = 60°

`∠`QOR 360° -( 90° + 90° + 60°)

`∠`QOR = 360° - 240°

`∠`QOR = 120°

ii) In `∠`QOR ,

OQ = QR (Radii of the same circle)

∴ OQR = `∠`QRO ………….(i)

But, `∠`OQR +`∠`QRO+ `∠`QOR =180°

`∠`OQR + `∠` QRO + 120° = 180°

`∠` OQR + `∠`QRO 60°

From (i)

2 `∠`OQR = 60°

`∠` OQR = 30°

iii) Now arc RQ subtends `∠`QOR at the centre and `∠`QSR at the remaining part of the circle.

∴ `∠` QSR = `1/2` `∠`QOR

⇒ `∠`QSR = `1/2xx120°`

⇒ `∠`QSR =60°

Is there an error in this question or solution?

Solution In the Following Figure, Pq and Pr Are Tangents to the Circle, with Centre O. If `∠`Qpr = 60°, Calculate: (I) ∠Qor (Ii) `∠`Oqr (Iii) `∠`Qsr 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.