MCQ
In Figure. $PB = PD.$ The value of $x$ is:
- A$85^\circ$
- B$90^\circ$
- ✓$25^\circ$
- D$35^\circ$
✓
Answer
Correct option: C.
$25^\circ$
$\text{In} \ \triangle\text{PBD},$ $\angle1+120^{\circ}=180^{\circ}$ $[\text{linear pair}]$
$\angle1=180^{\circ}-120^{\circ}=60^{\circ}$
Also, $\angle1=\angle2=60^{\circ}$ $[\because\text{PB}=\text{PD}]$
Also, $\angle2+\angle3=180^{\circ}$ $[\text{linear pair}]$
$\Rightarrow60^{\circ}+\angle3=180^{\circ} \ \Rightarrow \ \angle3=180^{\circ}-60^{\circ}$
$\Rightarrow\angle3=120^{\circ}$
$\text{In} \ \triangle\text{PDC},$ $35^{\circ}+\angle3+\text{x}^{\circ}=180^{\circ}$ [angle sum property of a triangle]
$\Rightarrow \ 35^{\circ}+120^{\circ}+\text{x}^{\circ}=180^{\circ}$
$\Rightarrow \ 155^{\circ}+\text{x}^{\circ}=180^{\circ}\Rightarrow \ \text{x}^{\circ}=180^{\circ}-155^{\circ}$
$\Rightarrow \ \text{x}=25^{\circ}$
$\angle1=180^{\circ}-120^{\circ}=60^{\circ}$
Also, $\angle1=\angle2=60^{\circ}$ $[\because\text{PB}=\text{PD}]$
Also, $\angle2+\angle3=180^{\circ}$ $[\text{linear pair}]$
$\Rightarrow60^{\circ}+\angle3=180^{\circ} \ \Rightarrow \ \angle3=180^{\circ}-60^{\circ}$
$\Rightarrow\angle3=120^{\circ}$
$\text{In} \ \triangle\text{PDC},$ $35^{\circ}+\angle3+\text{x}^{\circ}=180^{\circ}$ [angle sum property of a triangle]
$\Rightarrow \ 35^{\circ}+120^{\circ}+\text{x}^{\circ}=180^{\circ}$
$\Rightarrow \ 155^{\circ}+\text{x}^{\circ}=180^{\circ}\Rightarrow \ \text{x}^{\circ}=180^{\circ}-155^{\circ}$
$\Rightarrow \ \text{x}=25^{\circ}$
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