In the figure, $\angle AOF$ and $\angle FOG$ form a linear pair. $\angle EOB = \angle FOC = 90^o$ and $\angle DOC = \angle FOG = \angle AOB = 30^o$.Find the measures of $\angle FOE

Given:

$\angle AOF$ and $\angle FOG$ form a linear pair. $\angle EOB = \angle FOC = 90^o$ and $\angle DOC = \angle FOG = \angle AOB = 30^o$.

To do:

We have to find the measures of $\angle FOE, \angle COB$ and $\angle DOE$.

Solution:

We know that,

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

Therefore,

$\angle BOE + \angle AOB + \angle EOG = 180^o$

$30^o + 90^o + \angle EOG = 180^o$

$\angle EOG = 180^o - 30^o - 90^o$

$\angle EOG = 60^o$

$\angle FOG = 30^o$

$\Rightarrow \angle FOE = 60^o - 30^o = 30^o$

$\angle COD = 30^o, \angle COF = 90^o$

$\angle DOF = 90^o - 30^o$

$\angle DOF = 60^o$

$\angle DOE = \angle DOF - \angle EOF$

$\angle DOE = 60^o - 30^o = 30^o$

$\angle BOC = \angle BOE - \angle COE$

$\angle BOC = 90^o - 30^o - 30^o$

$\angle BOC = 90^o - 60^o = 30^o$

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Updated on: 10-Oct-2022

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