MCQ
In figure, for which value of $x$ is $I_1 \| l_2$ ?
- A$43$
- B$37$
- C$45$
- ✓$47$
✓
Answer
Correct option: D.
$47$
Let if $I_1 \| I_2$ and AB is tranverse to it
Then,
$\angle PBA$ should be equal to $\angle BAS ($Alternate angles$)$
So if $I_1 \| I_2$, then $\angle BAS =70^{\circ}$
$\Rightarrow \angle BAC=78^{\circ}-35^{\circ}=43^{\circ}\ldots (i)$
Now, in $\triangle ABC$
$x^0+\angle C+\angle B A C=180^{\circ}$
$\Rightarrow x^0+90^{\circ}+43^{\circ}=180^{\circ}$
$\Rightarrow x^0=180^{\circ}-90^{\circ}=43^{\circ}=47^{\circ}$
$\Rightarrow x^0=47^{\circ}$
So if $x^{\circ}=47^{\circ}$ then $I_1 \| I_2$
Then,
$\angle PBA$ should be equal to $\angle BAS ($Alternate angles$)$
So if $I_1 \| I_2$, then $\angle BAS =70^{\circ}$
$\Rightarrow \angle BAC=78^{\circ}-35^{\circ}=43^{\circ}\ldots (i)$
Now, in $\triangle ABC$
$x^0+\angle C+\angle B A C=180^{\circ}$
$\Rightarrow x^0+90^{\circ}+43^{\circ}=180^{\circ}$
$\Rightarrow x^0=180^{\circ}-90^{\circ}=43^{\circ}=47^{\circ}$
$\Rightarrow x^0=47^{\circ}$
So if $x^{\circ}=47^{\circ}$ then $I_1 \| I_2$
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