MCQ
- A120º
- B110º
- C80º
- D100º
✓
Answer
- 120º
Solution:
In $\triangle\text{ABC}$
$\angle\text{ABC}+\angle\text{BAC}+\angle\text{BCA}=180^\circ$ (Angle sum property)
$50^\circ+30^\circ+\angle\text{BCA}=180^\circ$
$\angle\text{BCA}=100^\circ$
In $\triangle\text{ECD}$
$\angle\text{ECD}+\angle\text{CDE}+\angle\text{CED}=180^\circ$ (Angle sum property)
$180^\circ-\angle\text{BCA}+40^\circ+\angle\text{CED}=180^\circ$
$\angle\text{CED}=100^\circ-40^\circ=60^\circ$
$\angle\text{CED}+\angle\text{AED}=180^\circ$ (Linear Pair)
$\angle\text{AED}=180^\circ-60^\circ=120^\circ.$
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