In the given figure, O is the centre of the circle and ∠AOC = 120°. Find ∠BDC.

45°

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30°

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15°

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60°

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Solution

The correct option is B 30°
Since BOA is the diameter,
$∠ A O C + ∠ B O C = 180 °$
$120 ° + ∠ B O C = 180 °$
$∠ B O C = 60 °$
$∠ B D C = \frac{1}{2} (∠ B O C)$

$∠ B D C = 30 °$
(The angle subtended by an arc of a circle at its centre is twice the angle it subtends anywhere on the circle’s circumference.)

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Similar questions

∠ A O C = 120^{\circ}

∠ B D C

In the given figure, O is the centre of a circle and $∠ A O C = 120^{\circ} .$ Then, $∠$ BDC = ?

(a) $60^{\circ}$

(b) $45^{\circ}$

(d) $15^{\circ}$

A O B

∠ A O C = 104^{o}

∠ B D C

A O B

∠ A O C = 110^{o}

∠ B D C

$In the given figure, O is the centre of a circle. If ∠ A O B = 40^{\circ} a n d ∠ B D C = 100^{\circ} t h e n f i n d ∠ O B C .$

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