Question
Find $x$ :
✓
Answer
Let us name the figure as following:
In $\triangle \mathrm{ABC}_r$
$\mathrm{AD}=\mathrm{AC} \ldots . . .[$ Given $]$
$\therefore \angle A D C=\angle A C D \quad......[$Angles opp. to equal sides are equal$]$
$\Rightarrow \angle \mathrm{ADC}=42^{\circ}$
Now,
$\angle \mathrm{ADC}=\angle \mathrm{DAB}+\angle \mathrm{DBA} \quad\dots... [$Exterior angle is equal to the sum of opp. interior angles$]$
But,
$\angle \mathrm{DAB}=\angle \mathrm{DBA}$
$[$Given: $\mathrm{BD}=\mathrm{DA}]$
$\therefore \angle \mathrm{ADC}=2 \angle \mathrm{DBA}$
$\Rightarrow 2 \angle \mathrm{DBA}=42^{\circ}$
$\Rightarrow \angle \mathrm{DBA}=21^{\circ}$
For $\mathrm{x}$ :
$\mathrm{x}=\angle \mathrm{CBA}+\angle \mathrm{BCA}[$Exterior angle is equal to the sum of opp. interior angles$]$
We know that,
$\angle C B A=21^{\circ}$
$\angle \mathrm{BCA}=42^{\circ}$
$\therefore x=21+42^{\circ}$
$\Rightarrow x=63^{\circ}$
In $\triangle \mathrm{ABC}_r$
$\mathrm{AD}=\mathrm{AC} \ldots . . .[$ Given $]$
$\therefore \angle A D C=\angle A C D \quad......[$Angles opp. to equal sides are equal$]$
$\Rightarrow \angle \mathrm{ADC}=42^{\circ}$
Now,
$\angle \mathrm{ADC}=\angle \mathrm{DAB}+\angle \mathrm{DBA} \quad\dots... [$Exterior angle is equal to the sum of opp. interior angles$]$
But,
$\angle \mathrm{DAB}=\angle \mathrm{DBA}$
$[$Given: $\mathrm{BD}=\mathrm{DA}]$
$\therefore \angle \mathrm{ADC}=2 \angle \mathrm{DBA}$
$\Rightarrow 2 \angle \mathrm{DBA}=42^{\circ}$
$\Rightarrow \angle \mathrm{DBA}=21^{\circ}$
For $\mathrm{x}$ :
$\mathrm{x}=\angle \mathrm{CBA}+\angle \mathrm{BCA}[$Exterior angle is equal to the sum of opp. interior angles$]$
We know that,
$\angle C B A=21^{\circ}$
$\angle \mathrm{BCA}=42^{\circ}$
$\therefore x=21+42^{\circ}$
$\Rightarrow x=63^{\circ}$
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