/_PQR =100 where P,Q,R,S are points on a circle with centre O find /_OPR.
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Hey there,
∠POR = 2 × ∠PQR = 2 × 100° = 200°
∴ ∠POR = 360° - 200° = 160°
In ΔOPR,
OP = OR (radii of the circle)
∠OPR = ∠ORP
Now,
∠OPR + ∠ORP +∠POR = 180° (Sum of the angles in a triangle)
⇒ ∠OPR + ∠OPR + 160° = 180°
⇒ 2∠OPR = 180° - 160°
⇒ ∠OPR = 10°
Hope this helps!
∠POR = 2 × ∠PQR = 2 × 100° = 200°
∴ ∠POR = 360° - 200° = 160°
In ΔOPR,
OP = OR (radii of the circle)
∠OPR = ∠ORP
Now,
∠OPR + ∠ORP +∠POR = 180° (Sum of the angles in a triangle)
⇒ ∠OPR + ∠OPR + 160° = 180°
⇒ 2∠OPR = 180° - 160°
⇒ ∠OPR = 10°
Hope this helps!
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