Question

In the given figure, $AB=AD$; $CB=CD$, $\angle A={42}^0$ and $\angle C={108}^0$, Find $\angle ABC$

• A. ${105}^0$
• B. ${150}^0$
• C. ${75}^0$
• D. ${110}^0$

Answer 1

The correct option is A ${105}^0$

Given: $AB=AD$; $CB=CD$, $\angle A={42}^0$ and $\angle C={108}^0$
Join AC.
Now, in $△ADC$ and $△ABC$
$AD=AB$ (Given)
$BC=CD$ (Given)
$AC=AC$ (Common)
Thus, $△ADC\cong △ABC$ (SSS rule)
Hence, $\angle CAB=\angle CAD=\frac{1}{2}\angle DAB=21$ (By CPCT)
$\angle ACB=\angle ACD=\frac{1}{2}\angle DCB=54$ (By CPCT)
Now, In $△ABC$
$\angle ABC+\angle CAB+\angle ACB=180$ (Angles sum property)
$\angle ABC+21+54=180$
$\angle ABC={105}^{\circ }$