Question
In the given figure, $AB\ ||\ CD$. Find the value of $x$.
✓
Answer
Since$ AB\ ||\ CD$ and AC is a transversal.
So, $\angle\text{BAC}+\angle\text{ACD}=180^\circ$ [sum of consecutive interior angles is $180^{\circ}$]
$\Rightarrow\angle\text{ACD}=180^\circ-\angle\text{BAC}$
$=180^\circ-75^\circ=105^\circ$
$\angle\text{ECF}=\angle\text{ACD}$ [Vertically opposite angles]
$\Rightarrow\angle\text{ECF}=105^\circ$
Now in $\triangle\text{CEF},$$\angle\text{ECF}+\angle\text{CEF}+\angle\text{EFC}=180^\circ$
$\Rightarrow105^\circ+\text{x}^\circ+30^\circ=180^\circ$
$\Rightarrow\text{x}=180-30-105=45$
Hence, $x = 45.$
So, $\angle\text{BAC}+\angle\text{ACD}=180^\circ$ [sum of consecutive interior angles is $180^{\circ}$]
$\Rightarrow\angle\text{ACD}=180^\circ-\angle\text{BAC}$
$=180^\circ-75^\circ=105^\circ$
$\angle\text{ECF}=\angle\text{ACD}$ [Vertically opposite angles]
$\Rightarrow\angle\text{ECF}=105^\circ$
Now in $\triangle\text{CEF},$$\angle\text{ECF}+\angle\text{CEF}+\angle\text{EFC}=180^\circ$
$\Rightarrow105^\circ+\text{x}^\circ+30^\circ=180^\circ$
$\Rightarrow\text{x}=180-30-105=45$
Hence, $x = 45.$
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