Question

In the following figure; $AC=CD$, $AD=BD$ and $\angle C={58}^{\circ }$. Find angle $CAB$.

• A. ${91.5}^o$
• B. ${82}^o$
• C. ${88.5}^o$
• D. none of the above

Answer 1

Answer: (A)

In $△ACD$,

by isosceles triangle property, angles opposite to equal sides are equal.

Then, let $\angle CAD=\angle CDA=x$.

We know, by angle sum property, the sum of angles $={180}^o$.
$\angle CAD+\angle CDA+\angle ACD={180}^o$

$⟹$$x+x+\angle ACD={180}^o$
$⟹$ $2x+{58}^o={180}^o$ ......(as $\angle C=\angle ACD={58}^o$)
$⟹$ $2x={122}^o$
$⟹$ $x={61}^{\circ }$.
Therefore, $\angle CAD=\angle CDA={61}^{\circ }$.