In the given figure- - COE -90- Find the value of x-A- 120-B- 60-C- 80-D- 30-

The correct option is D 30^∘Here, ∠BOD = ∠AOC nbsp; nbsp; nbsp;[Vertically opposite angles] ⇒ 2x = y Now, ∠AOC + ∠COE + ∠EOB = 180° ⇒ 2x + 90° + x = 1 ...

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Standard VI

Mathematics

Vertically Opposite Angles

In the given ...

Question

In the given figure, $∠ C O E = 90^{o}$ . Find the value of x.

120^{\circ}

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60^{\circ}

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80^{\circ}

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30^{\circ}

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Solution

The correct option is D $30^{\circ}$
Here, ∠BOD = ∠AOC [Vertically opposite angles]
⇒ 2x = y
Now, ∠AOC + ∠COE + ∠EOB = 180°
⇒ 2x + 90° + x = 180°
⇒ 3x = 90°
⇒ x = 30°

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In the given figure, ∠COE=90o. Find the value of x.

The correct option is D 30∘Here, ∠BOD = ∠AOC [Vertically opposite angles] ⇒ 2x = y Now, ∠AOC + ∠COE + ∠EOB = 180° ⇒ 2x + 90° + x = 180° ⇒ 3x = 90° ⇒ x = 30°

In the given figure, $∠ C O E = 90^{o}$ . Find the value of x.

The correct option is D $30^{\circ}$
Here, ∠BOD = ∠AOC [Vertically opposite angles]
⇒ 2x = y
Now, ∠AOC + ∠COE + ∠EOB = 180°
⇒ 2x + 90° + x = 180°
⇒ 3x = 90°
⇒ x = 30°