Question
✓
Answer
- 130º
Solution:
In $\triangle\text{ABC}$
$\angle\text{ABC}+\angle\text{ACB}+\angle\text{BAC}=180^\circ$ (Angle sum property)
$\angle\text{ACB}=180^\circ-25^\circ-45^\circ$
$\angle\text{ACB}=110^\circ$
$\angle\text{ACB}+\angle\text{ACD}=180^\circ$ (Linear pair)
$\angle\text{ACD}=180^\circ-110^\circ=70^\circ$
In $\triangle\text{CED}$
$\angle\text{AED}+\angle\text{EDC}+\angle\text{EDC}$ (Exterior angle is equal to sum of its two interior opposite angles)
$\angle\text{AED}=60^\circ+75^\circ=130^\circ.$
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