MCQ
In Fig, $PQ \| RS$, $\angle\text{AEF}=95^\circ,\angle\text{BHS}=110^\circ,$ and $\angle\text{ABC}=\text{x}^\circ.$ Then the value of $x$ is:
- A$70^\circ$
- B$15^\circ$
- ✓$25^\circ$
- D$35^\circ$
✓
Answer
Correct option: C.
$25^\circ$
Given that,
$PQ \| RS$
$\angle\text{AEF}=95^\circ$
$\angle\text{BHS}=110^\circ$
$\angle\text{ABC}=\text{x}^\circ$
$\angle\text{AEF}=\angle\text{AGH}=95^\circ$ (Corresponding angles)
$\angle\text{AGH}+\angle\text{HGB}=180^\circ$ (Linear pair)
$95^\circ+\angle\text{HGB}=180^\circ$
$\angle\text{HGB}=85^\circ$
$\angle\text{BHS}+\angle\text{BHG}=180^\circ $(Linear pair)
$110^\circ+\angle\text{BHG}=180^\circ$
$\angle\text{BHG}=70^\circ$
In $\triangle\text{BHG},$
$\angle\text{BHG}+\angle\text{HGB}+\angle\text{GBH}=180^\circ$
$70^\circ+85^\circ+\angle\text{GBH}=180^\circ$
$\angle\text{GBH}=25^\circ$
Thus,
$\angle\text{ABC}=\angle\text{GBH}=25^\circ.$
$PQ \| RS$
$\angle\text{AEF}=95^\circ$
$\angle\text{BHS}=110^\circ$
$\angle\text{ABC}=\text{x}^\circ$
$\angle\text{AEF}=\angle\text{AGH}=95^\circ$ (Corresponding angles)
$\angle\text{AGH}+\angle\text{HGB}=180^\circ$ (Linear pair)
$95^\circ+\angle\text{HGB}=180^\circ$
$\angle\text{HGB}=85^\circ$
$\angle\text{BHS}+\angle\text{BHG}=180^\circ $(Linear pair)
$110^\circ+\angle\text{BHG}=180^\circ$
$\angle\text{BHG}=70^\circ$
In $\triangle\text{BHG},$
$\angle\text{BHG}+\angle\text{HGB}+\angle\text{GBH}=180^\circ$
$70^\circ+85^\circ+\angle\text{GBH}=180^\circ$
$\angle\text{GBH}=25^\circ$
Thus,
$\angle\text{ABC}=\angle\text{GBH}=25^\circ.$
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