In the diagram above, if $∠ B O E$ is twice the $∠ D O E$ , then $∠ B O E =$

72^{\circ}

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36^{\circ}

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64^{\circ}

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Solution

The correct option is A $72^{\circ}$
Given $∠ B O E = 2 \times ∠ D O E$
Let $∠ D O E = x \Rightarrow ∠ B O E = 2 x$ .
Also, $∠ A O C = 108^{\circ}$ .
$∠ A O C$ and $∠ D O B$ are vertical angles
$\Rightarrow ∠ A O C = ∠ D O B = 108^{\circ}$
$\Rightarrow ∠ D O E + ∠ B O E = 108^{\circ}$
$\Rightarrow x + 2 x = 108^{\circ} \Rightarrow 3 x = 108^{\circ}$
$\Rightarrow x = 36^{\circ}$
$∴ ∠ B O E = 2 x = 2 \times 36^{\circ} = 72^{\circ}$

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In the diagram above, if ∠BOE is twice the ∠DOE, then ∠BOE=

The correct option is A 72∘Given ∠BOE=2×∠DOE Let ∠DOE=x⇒∠BOE=2x. Also, ∠AOC=108∘. ∠AOC and ∠DOB are vertical angles ⇒∠AOC=∠DOB=108∘ ⇒∠DOE+∠BOE=108∘ ⇒x+2x=108∘⇒3x=108∘ ⇒x=36∘ ∴∠BOE=2x=2×36∘=72∘

In the diagram above, if $∠ B O E$ is twice the $∠ D O E$ , then $∠ B O E =$