A circle with centre O is given in which ∠OBA = 30° and ∠OCA = 40°. Find ∠BOC.

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Solution

We have:
OA = OB (Radii of a circle)
⇒ ∠OAB = ∠OBA = 30°
Similarly, OA = OC (Radii of a circle)
⇒ ∠OAC = ∠OCA = 40°
∴ ∠BAC = ∠OAB + ∠OAC
= (30° + 40°) = 70°
Now, ∠BOC = 2∠BAC
= (2 × 70°) = 140°
∴ ∠BOC = 140°

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A circle with centre O is given in which ∠OBA = 30° and ∠OCA = 40°. Find ∠BOC.

We have: OA = OB (Radii of a circle) ⇒ ∠OAB = ∠OBA = 30° Similarly, OA = OC (Radii of a circle) ⇒ ∠OAC = ∠OCA = 40° ∴ ∠BAC = ∠OAB + ∠OAC = (30° + 40°) = 70° Now, ∠BOC = 2∠BAC = (2 × 70°) = 140° ∴ ∠BOC = 140°

We have:
OA = OB (Radii of a circle)
⇒ ∠OAB = ∠OBA = 30°
Similarly, OA = OC (Radii of a circle)
⇒ ∠OAC = ∠OCA = 40°
∴ ∠BAC = ∠OAB + ∠OAC
= (30° + 40°) = 70°
Now, ∠BOC = 2∠BAC
= (2 × 70°) = 140°
∴ ∠BOC = 140°