In the figure, lines $AB, CD$ and $EF$ intersect at $O$. Find the measures of $\angle AOC, \angle COF, \angle DOE$ and $\angle BOF$."

Given:

Lines $AB, CD$ and $EF$ intersect at $O$.

To do:

We have to find the measures of $\angle AOC, \angle COF, \angle DOE$ and $\angle BOF$.

Solution:

We know that,

Vertically opposite angles are equal.

Sum of the angles on a straight line is $180^o$.

Therefore,

$\angle AOC=\angle BOD=35^o$       (Vertically opposite angles)

$\angle BOF=\angle AOE=40^o$       (Vertically opposite angles)

$AOB$ is a straight line.

This implies,

$\angle AOE+\angle EOD+\angle BOD = 180^o$

$40^o + \angle EOD + 35^o = 180^o$

$\angle EOD= 180^o-75^o$

$\angle EOD=105^o$

$\angle COF=\angle EOD=105^o$       (Vertically opposite angles)

Hence, $\angle AOC=35^o, \angle COF=105^o, \angle DOE=105^o$ and $\angle BOF=40^o$.

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Updated on: 10-Oct-2022

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