Question
In Fig. for which value of $x$ is $l_1 || l_2$?
✓
Answer
Let if$ l_1 || l_2$ and$ AB$ is tranverse to it.
Then,
$\angle\text{PBA}$ should be equal $\angle\text{BAS} ($Alternate angles$)$
So if $l_1 || l_2,$ then $\angle\text{BAS}=70^\circ$
$\Rightarrow\angle\text{BAC}=78^\circ-35^\circ=43^\circ\dots(1)$
Now, in $\triangle\text{ABC}$
$\text{x}^\circ+\angle\text{C}+\angle\text{BAC}=180^\circ$
$\Rightarrow\text{x}^\circ+90^\circ+43^\circ=180^\circ$
$\Rightarrow\text{x}^\circ=180^\circ-90^\circ-43^\circ=47^\circ$
$\Rightarrow\text{x}^\circ=47^\circ$
So if $x^\circ = 47^\circ$ then $l_1 || l_2$
Then,
$\angle\text{PBA}$ should be equal $\angle\text{BAS} ($Alternate angles$)$
So if $l_1 || l_2,$ then $\angle\text{BAS}=70^\circ$
$\Rightarrow\angle\text{BAC}=78^\circ-35^\circ=43^\circ\dots(1)$
Now, in $\triangle\text{ABC}$
$\text{x}^\circ+\angle\text{C}+\angle\text{BAC}=180^\circ$
$\Rightarrow\text{x}^\circ+90^\circ+43^\circ=180^\circ$
$\Rightarrow\text{x}^\circ=180^\circ-90^\circ-43^\circ=47^\circ$
$\Rightarrow\text{x}^\circ=47^\circ$
So if $x^\circ = 47^\circ$ then $l_1 || l_2$
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