Question
In the figure given alongside, find:
- $\angle\text{ACD}$
- $\angle\text{AED}$
✓
Answer
In $\triangle\text{ABC},$ side BC is produced to D From D, draw a line meeting AC at E, so that $\angle\text{D}=40^\circ$.
$\angle\text{A}=25^\circ,\angle\text{B}=45^\circ$
Hence, $\angle\text{ACD}=70^\circ$ and $\angle\text{AED}=110^\circ$
$\angle\text{A}=25^\circ,\angle\text{B}=45^\circ$
- In $\triangle\text{ABC},$
- In $\triangle\text{CDE},$
Hence, $\angle\text{ACD}=70^\circ$ and $\angle\text{AED}=110^\circ$
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