Question
✓
Answer
- 75º
Solution:
$\angle\text{QPR}=\angle\text{PRT}=40^\circ$ (Alternate interior angles)
In $\triangle\text{QPR}$
$\angle\text{PQR}+\angle\text{QPR}+\angle\text{PRQ}=180\circ $ (Angle sum property)
$65^\circ+40^\circ+\text{x}^\circ=180^\circ$
$\text{x}^{\circ}=180^\circ-40^\circ-65^\circ$
$\text{x}^{\circ}=75$
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