In the given figure, O is the centre of a circle and ∠OAB = 50°. Then, ∠BOD = ?
(a) 130°
(b) 50°
(c) 100°
(d) 80°

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Solution

(c) 100°
OA = OB (Radii of a circle)
⇒ ∠OBA = ∠OAB = 50°
In Δ OAB, we have:
∠ OAB + ∠OBA + ∠AOB = 180° (Angle sum property of a triangle)
⇒ 50° + 50° + ∠AOB = 180°
⇒ ∠AOB = (180° - 100°) = 80°
Since ∠AOB + ∠BOD = 180° (Linear pair)
∴ ∠BOD = (180° - 80°) = 100°

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Similar questions

In the given figure, O is the centre of a circle and $∠ O A B = 50^{\circ}, T h e n, ∠ B O D = ?$

(a) $130^{\circ}$

(b) $50^{\circ}$

(d) $80^{\circ}$

In the given figure, O is the centre of a circle. If $∠ O A B = 40^{\circ}$ and C is a point on the circle then $∠ A C B = ?$

(a) $40^{\circ}$

(b) $50^{\circ}$

(d) $100^{\circ}$

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