Question
Calculate $x$ :
✓
Answer
Let the triangle be $\text{ABC}$ and the altitude be $A D$.
In $\triangle \mathrm{ABD}$, $\angle \mathrm{DBA}=\angle \mathrm{DAB}=37^{\circ} \ldots . .[$ Given $\mathrm{BD}=\mathrm{AD}$ and angles opposite to equal sides are equal$]$
Now, $\angle \mathrm{CDA}=\angle \mathrm{DBA}+\angle \mathrm{DAB} \ldots \ldots. [$Exterior angle is equal to the sum of opp. interior angles$]$
$\therefore \angle C D A=37^{\circ}+37^{\circ}$
$ \Rightarrow \angle C D A=74^{\circ}$
Now in $\triangle A D C$,
$\angle C D A=\angle C A D=74^{\circ} \ldots .[$ Given $\mathrm{CD}=\mathrm{AC}$ and angles opposite to equal sides are equal$]$
$ \angle C A D+\angle C D A+\angle A C D=180^{\circ}$
$ \Rightarrow 74^{\circ}+74^{\circ}+x=180^{\circ}$
$ \Rightarrow x=180^{\circ}-148^{\circ}$
$ \Rightarrow x=32^{\circ}$
In $\triangle \mathrm{ABD}$, $\angle \mathrm{DBA}=\angle \mathrm{DAB}=37^{\circ} \ldots . .[$ Given $\mathrm{BD}=\mathrm{AD}$ and angles opposite to equal sides are equal$]$
Now, $\angle \mathrm{CDA}=\angle \mathrm{DBA}+\angle \mathrm{DAB} \ldots \ldots. [$Exterior angle is equal to the sum of opp. interior angles$]$
$\therefore \angle C D A=37^{\circ}+37^{\circ}$
$ \Rightarrow \angle C D A=74^{\circ}$
Now in $\triangle A D C$,
$\angle C D A=\angle C A D=74^{\circ} \ldots .[$ Given $\mathrm{CD}=\mathrm{AC}$ and angles opposite to equal sides are equal$]$
$ \angle C A D+\angle C D A+\angle A C D=180^{\circ}$
$ \Rightarrow 74^{\circ}+74^{\circ}+x=180^{\circ}$
$ \Rightarrow x=180^{\circ}-148^{\circ}$
$ \Rightarrow x=32^{\circ}$
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