Question
In Fig. $AB || CD, AF || ED,$ $\angle\text{AFC} = 68^\circ$ and $\angle\text{FED} = 42^\circ$. Find $\angle\text{FED} = 42^\circ$.
✓
Answer
$AF$ and $ED$ are parallel and $EF$ is transversal.
Then, $\angle\text{AFE}=\angle\text{FED} [$Alternate interior angls$]$
$\Rightarrow\angle\text{AFE}=42^\circ$
$\because\angle\text{FED=42}^\circ$
Now, $\angle\text{AFC}+\angle\text{AFE}+\angle\text{EFD}=180^\circ [ \because$ sum of all the angles on a straight line is $180^\circ ] $
$\Rightarrow68^\circ+42^\circ+\text{EFD}=180^\circ$
$\Rightarrow110^\circ+\angle\text{EFD}=180^\circ$
$\Rightarrow\text{EFD}=180^\circ-110^\circ=70^\circ$
Then, $\angle\text{AFE}=\angle\text{FED} [$Alternate interior angls$]$
$\Rightarrow\angle\text{AFE}=42^\circ$
$\because\angle\text{FED=42}^\circ$
Now, $\angle\text{AFC}+\angle\text{AFE}+\angle\text{EFD}=180^\circ [ \because$ sum of all the angles on a straight line is $180^\circ ] $
$\Rightarrow68^\circ+42^\circ+\text{EFD}=180^\circ$
$\Rightarrow110^\circ+\angle\text{EFD}=180^\circ$
$\Rightarrow\text{EFD}=180^\circ-110^\circ=70^\circ$
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