MCQ
In Fig. if $AB \| CD$, then $x =$
- A$105^\circ$
- B$115^\circ$
- ✓$100^\circ$
- D$110^\circ$
✓
Answer
Correct option: C.
$100^\circ$
Given that,$AB \| CD$
Produce $P$ to $Q$ so that $PQ \| AB \| CD$
$\angle\text{BAP}+\angle\text{APQ}=180^\circ$ (Interior angle)
$132^\circ+\angle\text{APQ}=180^\circ$
$\angle\text{APQ}=48^\circ\text{(i)}$
$\angle\text{APC}=\angle\text{APQ}+\angle\text{QPC}$
$148^\circ=48^\circ+\angle\text{QPC}$ [From (i)]
$\angle\text{QPC}=100^\circ$
$\angle\text{QPC}+\angle\text{PCD}=180^\circ$ (Interior angles)
$100^\circ+\angle\text{PCD}=180^\circ$
$\angle\text{PCD}=80^\circ$
$\angle\text{PCD}+\text{x}=180^\circ$(Linear pair)
$80^\circ+\text{x}=180^\circ$
$\text{x}=100^\circ.$
Produce $P$ to $Q$ so that $PQ \| AB \| CD$
$\angle\text{BAP}+\angle\text{APQ}=180^\circ$ (Interior angle)
$132^\circ+\angle\text{APQ}=180^\circ$
$\angle\text{APQ}=48^\circ\text{(i)}$
$\angle\text{APC}=\angle\text{APQ}+\angle\text{QPC}$
$148^\circ=48^\circ+\angle\text{QPC}$ [From (i)]
$\angle\text{QPC}=100^\circ$
$\angle\text{QPC}+\angle\text{PCD}=180^\circ$ (Interior angles)
$100^\circ+\angle\text{PCD}=180^\circ$
$\angle\text{PCD}=80^\circ$
$\angle\text{PCD}+\text{x}=180^\circ$(Linear pair)
$80^\circ+\text{x}=180^\circ$
$\text{x}=100^\circ.$
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