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

The correct option is A ${91.5}^o$

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, $\angle CAD=\angle CDA={61}^{\circ }$.
Now, in $△ADB$,
$AD=BD$ (Given)
$\angle DAB=\angle ABD=y$ ......(Isosceles triangle property).
Also, $\angle CDA=\angle DBA+\angle DAB$ ......(Exterior angle property)
$⟹$ ${61}^o=y+y$
$⟹$ $2y={61}^o$
$⟹$ $y={30.5}^0$
Then, $\angle CAB=\angle CAD+\angle DAB$
$⟹$ $\angle CAB={61}^o+{30.5}^o$
$⟹$ $\angle CAB={91.5}^{\circ }$.

Therefore, option $A$ is correct.