Line DE intersects PQ at O- If - POE - 3Y- - QOE - 2X and X - 30- then 6Y - 3X-

Given, x = 30^∘, so 2x = 60^∘ Also, 2x + 3y = 180^∘ (linear pair) Replacing x in the above equation, 60^∘ + 3y = 180^∘ 3y = 120^∘ y = 40^∘ The value of ( ...

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Standard IX

Mathematics

Angles and Types of Angles

Line DE inter...

Question

Line DE intersects PQ at O. If $∠ P O E = 3 Y, ∠ Q O E = 2 X$ and X = $30^{\circ}$ , then (6Y - 3X)=

60^{\circ}

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150^{\circ}

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180^{\circ}

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Solution

The correct option is B $150^{\circ}$

Given, x = $30^{\circ}$ , so 2x = $60^{\circ}$

Also, 2x + 3y = $180^{\circ}$ (linear pair)

Replacing x in the above equation,
$60^{\circ}$ + 3y = $180^{\circ}$

3y = $120^{\circ}$

y = $40^{\circ}$

The value of (6y - 3x) = $150^{\circ}$

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Line DE intersects PQ at O. If ∠POE=3Y,∠QOE=2X and X = 30∘, then (6Y - 3X)=

The correct option is B 150∘ Given, x = 30∘, so 2x = 60∘ Also, 2x + 3y = 180∘ (linear pair) Replacing x in the above equation, 60∘ + 3y = 180∘ 3y = 120∘ y = 40∘ The value of (6y - 3x) = 150∘

Line DE intersects PQ at O. If $∠ P O E = 3 Y, ∠ Q O E = 2 X$ and X = $30^{\circ}$ , then (6Y - 3X)=

The correct option is B $150^{\circ}$

Given, x = $30^{\circ}$ , so 2x = $60^{\circ}$

Also, 2x + 3y = $180^{\circ}$ (linear pair)

Replacing x in the above equation,
$60^{\circ}$ + 3y = $180^{\circ}$

3y = $120^{\circ}$

y = $40^{\circ}$

The value of (6y - 3x) = $150^{\circ}$