Question
In the figure given alongside, find:
$i. \angle\text{ACD}$
$ii. \angle\text{AED}$
$i. \angle\text{ACD}$
$ii. \angle\text{AED}$
✓
Answer
In $\triangle\text{ABC},$ side $BC$ is produced to $D$ From $D,$ draw a line meeting $AC$ at $E,$ so that $\angle\text{D}=40^\circ$. $\angle\text{A}=25^\circ,\angle\text{B}=45^\circ$
$i.$ In $\triangle\text{ABC},$
Exterior $\angle\text{ACD} = \angle\text{A}+\angle\text{B}=25^\circ+45^\circ=70^\circ$
$ii.$ In $\triangle\text{CDE},$
Exterior $\angle\text{AED} = \angle\text{ECD}+\angle\text{D}= \angle\text{ACD}+\angle\text{D}=70^\circ+40^\circ=7=110^\circ$
Hence, $\angle\text{ACD}=70^\circ$ and $\angle\text{AED}=110^\circ$
$i.$ In $\triangle\text{ABC},$
Exterior $\angle\text{ACD} = \angle\text{A}+\angle\text{B}=25^\circ+45^\circ=70^\circ$
$ii.$ In $\triangle\text{CDE},$
Exterior $\angle\text{AED} = \angle\text{ECD}+\angle\text{D}= \angle\text{ACD}+\angle\text{D}=70^\circ+40^\circ=7=110^\circ$
Hence, $\angle\text{ACD}=70^\circ$ and $\angle\text{AED}=110^\circ$
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