Question
Calculate $x$ :
✓
Answer
Let triangle be $\text{ABC}$ and altitude be $A D$.
In $\triangle \mathrm{ABD}$, $\angle \mathrm{DBA}=\angle \mathrm{DAB}=50^{\circ} \ldots[$ Given $\mathrm{BD}=\mathrm{AD}$ and angles opposite to equal sides are equal$]$
Now,
$\angle \mathrm{CDA}=\angle \mathrm{DBA}+\angle \mathrm{DAB} \ldots [$Exterior angle is equal to the sum of opp. interior angles$]$
$\therefore \angle C D A=50^{\circ}+50^{\circ}$
$ \Rightarrow \angle C D A=100^{\circ}$
In $\triangle \mathrm{ADC}$
$\angle \mathrm{DAC}=\angle \mathrm{DCA}=\mathrm{x} \ldots [$Given $\mathrm{AD}=\mathrm{DC}$ and angles opposite to equal sides are equal$]$
$\therefore \angle \mathrm{DAC}+\angle \mathrm{DCA}+\angle \mathrm{ADC}=180^{\circ}$
$ \Rightarrow \mathrm{x}+\mathrm{x}+100^{\circ}=180^{\circ}$
$ \Rightarrow 2 \mathrm{x}=80^{\circ}$
$ \Rightarrow \mathrm{x}=40^{\circ}$
In $\triangle \mathrm{ABD}$, $\angle \mathrm{DBA}=\angle \mathrm{DAB}=50^{\circ} \ldots[$ Given $\mathrm{BD}=\mathrm{AD}$ and angles opposite to equal sides are equal$]$
Now,
$\angle \mathrm{CDA}=\angle \mathrm{DBA}+\angle \mathrm{DAB} \ldots [$Exterior angle is equal to the sum of opp. interior angles$]$
$\therefore \angle C D A=50^{\circ}+50^{\circ}$
$ \Rightarrow \angle C D A=100^{\circ}$
In $\triangle \mathrm{ADC}$
$\angle \mathrm{DAC}=\angle \mathrm{DCA}=\mathrm{x} \ldots [$Given $\mathrm{AD}=\mathrm{DC}$ and angles opposite to equal sides are equal$]$
$\therefore \angle \mathrm{DAC}+\angle \mathrm{DCA}+\angle \mathrm{ADC}=180^{\circ}$
$ \Rightarrow \mathrm{x}+\mathrm{x}+100^{\circ}=180^{\circ}$
$ \Rightarrow 2 \mathrm{x}=80^{\circ}$
$ \Rightarrow \mathrm{x}=40^{\circ}$
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