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) $50^{\circ}$

(c) $100^{\circ}$

(d) $80^{\circ}$

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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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