MCQ
In Fig. for which value of x is $I_1 \| I_2$?
- A$37^{\circ}$
- B$43^{\circ}$
- C$45^{\circ}$
- ✓$47^{\circ}$
✓
Answer
Correct option: D.
$47^{\circ}$
Let if $I_1 \| I_2$ and $AB$ is tranverse to it.
Then,
$\angle\text{PBA}$ should be equal $\angle\text{BAS}$ (Alternate angles)
So if $I_1 \| I_2$, 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 $I_1 \| I_2$
Then,
$\angle\text{PBA}$ should be equal $\angle\text{BAS}$ (Alternate angles)
So if $I_1 \| I_2$, 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 $I_1 \| I_2$
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