Question

As shown in the figure $AC=CD,\angle CAB-\angle ABC={30}^{\circ }$. Then $\angle BAD$ has a measure of:

• A. $15{}^{\circ }$
• B. $30{}^{\circ }$
• C. $20{}^{\circ }$
• D. ${\left(22\frac{1}{2}\right)}^{\circ }$

Answer 1

The correct option is A $15{}^{\circ }$

Given, $AC=CD$
$\angle CAD=\angle CDA$ .......... $\left(i\right)$ (Isosceles triangle property)
Also, $\angle CDA=\angle DAB+\angle ABC$
Given, $\angle CAB-\angle ABC={30}^{\circ }$
$\angle CAD+\angle BAD-(\angle CDA-\angle DAB)={30}^{\circ }$
$\angle CAD-\angle CDA+2\angle BAD={30}^{\circ }$
$\Rightarrow \angle BAD={15}^{\circ }$ .... Using $\left(i\right)$