Question
In the figure given alongside, find:
$i. \angle\text{ACD}$
$ii. \angle\text{ADC}$
$iii. \angle\text{DAE}$
$i. \angle\text{ACD}$
$ii. \angle\text{ADC}$
$iii. \angle\text{DAE}$
✓
Answer
$i.$ In $\triangle\text{ABC},$ sides $BC$ is produced to $D$ and $BA$ to $E.$
$\angle\text{CAD} = 50^\circ,\angle\text{B} = 40^\circ$ and $\angle\text{ACB} = 100^\circ$
$\angle\text{ACB} +\angle\text{ACD} = 180^\circ ($Linear pair$)$
$\Rightarrow100^\circ+\angle\text{ACD} = 180^\circ$
$\Rightarrow\angle\text{ACD} = 180^\circ-100^\circ=80^\circ$
$ii.$ In $\triangle\text{ABC},$
$\angle\text{CAD} +\angle\text{ACD}++\angle\text{ADC} = 180^\circ ($sum of angles of a triangle$)$
$\Rightarrow50^\circ+80^\circ+\angle\text{ADC} = 180^\circ$
$\Rightarrow130^\circ+\angle\text{ADC}=180^\circ$
$\Rightarrow\angle\text{ADC}=180^\circ-130^\circ=50^\circ$
$iii.$ Now, in $\triangle\text{ABD}, BA$ is produced to $E.$
Exterior $\angle\text{DAE} =\angle\text{ACD} +\angle\text{ADC}=80^\circ+50^\circ=130^\circ$
Hence, $\angle\text{ACD}=80^\circ,\angle\text{ADC}=50^\circ$and $\angle\text{DAE}=130^\circ$
$\angle\text{CAD} = 50^\circ,\angle\text{B} = 40^\circ$ and $\angle\text{ACB} = 100^\circ$
$\angle\text{ACB} +\angle\text{ACD} = 180^\circ ($Linear pair$)$
$\Rightarrow100^\circ+\angle\text{ACD} = 180^\circ$
$\Rightarrow\angle\text{ACD} = 180^\circ-100^\circ=80^\circ$
$ii.$ In $\triangle\text{ABC},$
$\angle\text{CAD} +\angle\text{ACD}++\angle\text{ADC} = 180^\circ ($sum of angles of a triangle$)$
$\Rightarrow50^\circ+80^\circ+\angle\text{ADC} = 180^\circ$
$\Rightarrow130^\circ+\angle\text{ADC}=180^\circ$
$\Rightarrow\angle\text{ADC}=180^\circ-130^\circ=50^\circ$
$iii.$ Now, in $\triangle\text{ABD}, BA$ is produced to $E.$
Exterior $\angle\text{DAE} =\angle\text{ACD} +\angle\text{ADC}=80^\circ+50^\circ=130^\circ$
Hence, $\angle\text{ACD}=80^\circ,\angle\text{ADC}=50^\circ$and $\angle\text{DAE}=130^\circ$
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