Question

Calculate the value of angle $x$ (in degrees) in the above figure:

Answer 1

Given: $AD=CD,AC=BC,\angle ABC={37}^{\circ }$

To find: $x^{\circ }$

Proof: In $△ACD$

$AD=CD$[Given}

$\Rightarrow \angle DAC=\angle DCA=y^{\circ }\left(say\right)$[Angles opposite to equal side are equal]

Similarly, In $△ABC$

$AC=BC$[Given]

$\Rightarrow \angle CAB=\angle ABC={37}^{\circ }$[Angles opposite to equal sides are equal]

Now, By angle sum property in $△ABC$, we have

$\angle CAB+\angle ABC+\angle ACB={180}^{\circ }$

$\Rightarrow {37}^{\circ }+{37}^{\circ }+\angle ACB={180}^{\circ }$

$\Rightarrow \angle ACB={106}^{\circ }$

Again, By angle sum property in $△ACD$, we have

$\angle DAC+\angle DCA+\angle ADC={180}^{\circ }$
$\Rightarrow y^{\circ }+y^{\circ }+x^{\circ }={180}^{\circ }$

$\Rightarrow y^{\circ }=\frac{{180}^{\circ }-x^{\circ }}{2}...\left(1\right)$

Now, By Exterior angle sum property, we have

$\angle ACB=\angle ADC+\angle DAC$

$\Rightarrow {106}^{\circ }=x^{\circ }+y^{\circ }$

$\Rightarrow {106}^{\circ }=x^{\circ }+\frac{{180}^{\circ }-x^{\circ }}{2}$[using $\left(1\right)$]

$\Rightarrow \frac{2x^{\circ }+{180}^{\circ }-x^{\circ }}{2}={106}^{\circ }$

$\Rightarrow x^{\circ }+{180}^{\circ }={212}^{\circ }$

$\Rightarrow x^{\circ }={32}^{\circ }$