In-the-given-figure- O-is-the-centre-of a circle- If - AOB - 40-and- BDC - 100-then -find- OBC- 60-80-120-40-

The correct option is A 60^∘ ∠ AOB=2×∠ ACB ⇒∠ ACB=1/2×∠ AOB =1/2× 40^∘=20^∘ In Δ BDC,∠ DBC+∠ BDC+∠ DCB=180^∘ ⇒∠ DBC+100^∘+20^∘=180^∘ ∴∠ DBC=180^∘ 120^∘=60^∘ ...

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Byju's Answer

Standard X

Mathematics

Arc/Chord Properties of Circles

In the given ...

Question

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

$60^{\circ}$

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$40^{\circ}$

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$80^{\circ}$

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$120^{\circ}$

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Solution

The correct option is A
$60^{\circ}$

$∠ A O B = 2 \times ∠ A C B$
$\Rightarrow ∠ A C B = \frac{1}{2} \times ∠ A O B$
$= \frac{1}{2} \times 40^{\circ} = 20^{\circ}$

In $Δ B D C, ∠ D B C + ∠ B D C + ∠ D C B = 180^{\circ}$
$\Rightarrow ∠ D B C + 100^{\circ} + 20^{\circ} = 180^{\circ}$
$∴ ∠ D B C = 180^{\circ} - 120^{\circ} = 60^{\circ}$

$H e n c e ∠ O B C = 60^{\circ}$

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In the given figure, O is the centre of a circle. If ∠AOB=40∘ and ∠BDC=100∘ then find ∠OBC.

The correct option is A 60∘ ∠AOB=2×∠ACB ⇒∠ACB=12×∠AOB =12×40∘=20∘ In ΔBDC,∠DBC+∠BDC+∠DCB=180∘ ⇒∠DBC+100∘+20∘=180∘ ∴∠DBC=180∘−120∘=60∘ Hence ∠OBC=60∘

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