Three lines AB- CD and EF interesect each other at O- If - AOE -30- and - DOB -40- see the figure- Find - COF-110-70-55-n 90-

The correct option is A 110^∘ From the given figure, we have ∠ AOE + ∠ EOD + ∠ DOB = 180^∘ [∴Sum of all ...

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Standard VI

Mathematics

Finding Unknown Angles

Three lines A...

Question

Three lines AB, CD and EF interesect each other at O. If $∠ A O E = 30^{\circ} a n d ∠ D O B = 40^{\circ}$ (see the figure). Find $∠ C O F .$

$70^{\circ}$

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$110^{\circ}$

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$55^{\circ}$

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$90^{\circ}$

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Solution

The correct option is B

$110^{\circ}$

From the given figure, we have
$∠ A O E + ∠ E O D + ∠ D O B = 180^{\circ} [∴ S u m o f a l l a n g l e s o n a s t r a i g h t l i n e i s 180^{\circ}] \Rightarrow 30^{\circ} + ∠ E O D + 40^{\circ} = 180^{\circ} \Rightarrow ∠ E O D = 180^{\circ} - 70^{\circ} \Rightarrow ∠ E O D = 110^{\circ}$
Again, $∠ E O D = ∠ C O F [v e r t i c a l l y o p p o s i t e a n g l e s] \Rightarrow ∠ C O F = 110^{\circ}$

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Three lines AB, CD and EF interesect each other at O. If ∠AOE=30∘ and ∠DOB=40∘ (see the figure). Find ∠COF.

The correct option is B 110∘ From the given figure, we have ∠AOE+∠EOD+∠DOB=180∘ [∴Sum of all angles on a straight line is 180∘]⇒30∘+∠EOD+40∘=180∘⇒∠EOD=180∘−70∘⇒∠EOD=110∘ Again, ∠EOD=∠COF [vertically opposite angles]⇒∠COF=110∘

Three lines AB, CD and EF interesect each other at O. If $∠ A O E = 30^{\circ} a n d ∠ D O B = 40^{\circ}$ (see the figure). Find $∠ C O F .$