Given, PQ and PR are two tangents to a circle with centre O and ∠QPR = 46°.
Since, the radius drawn to these tangents will be perpendicular to the tangents,
OQ ⊥ PQ and OR ⊥ PR.
∴ ∠OQP = ∠ORP = 90°
We know that the sum of angles in a quadrilateral is 360°.
⇒ ∠OQP + ∠QPR + ∠ORP + ∠QOR = 360°
⇒ 90° + 46° + 90° + ∠QOR = 360°
⇒ 226° + ∠QOR = 360°
⇒ ∠QOR = 360° - 226°
∴ ∠QOR = 134°
Ans. ∠QOR = 134°
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