In the given figure- chords AB-CD- - COD-60- and - AOD-170-Find the value of - BOC-

We know that equal chords of a circle subtend equal angles nbsp;at the centre. ⇒∠ AOB=∠ COD=60^∘ Since, the central angle of a circle is 360^∘. So, ∠ AOB+∠ ...

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Standard X

Mathematics

Extension of Inscribed Angle Theorem

In the given ...

Question

In the given figure, chords $A B = C D, ∠ C O D = 60^{\circ}$ , and $∠ A O D = 170^{\circ}$ .

Find the value of $∠ B O C$ .

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Solution

We know that equal chords of a circle subtend equal angles at the centre.

$\Rightarrow ∠ A O B = ∠ C O D = 60^{\circ}$

Since, the central angle of a circle is $360^{\circ}$ .

So, $∠ A O B + ∠ B O C + ∠ C O D + ∠ A O D = 360^{\circ}$

$\Rightarrow 60^{\circ} + ∠ B O C + 60^{\circ} + 170^{\circ} = 360^{\circ}$

$\Rightarrow 290^{\circ} + ∠ B O C = 360^{\circ}$

$\Rightarrow ∠ B O C = 360^{\circ} - 290^{\circ} = 70^{\circ}$

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In the given figure, chords AB=CD, ∠COD=60∘, and ∠AOD=170∘. Find the value of ∠BOC.

We know that equal chords of a circle subtend equal angles at the centre. ⇒∠AOB=∠COD=60∘ Since, the central angle of a circle is 360∘. So, ∠AOB+∠BOC+∠COD+∠AOD=360∘ ⇒60∘ +∠BOC+60∘ +170∘ =360∘ ⇒290∘ +∠BOC=360∘ ⇒∠BOC=360∘ −290∘ =70∘

In the given figure, chords $A B = C D, ∠ C O D = 60^{\circ}$ , and $∠ A O D = 170^{\circ}$ .

Find the value of $∠ B O C$ .