Question

In the following figure: $AC=CD,AD=BD$ and $\angle C={58}^{\circ }$. Find $\angle$ $CAB$(in degrees) to nearest integer.

Answer 1

In $△ACD$,
$\angle CAD=\angle CDA=x$
Sum of angles = 180
$\angle CAD+\angle CDA+\angle ACD=180$
$2x+58=180$
$2x=122$
$x={61}^{\circ }$
$\angle CAD=\angle CDA={61}^{\circ }$
In $△ADB$,
$AD=BD$ (Given)
$\angle DAB=\angle ABD=y$ (Isosceles triangle property)
$\angle CDA=\angle DBA+\angle DAB$
$61=y+y$
$2y=61$
$y=30.5$
Now, $\angle CAB=\angle CAD+\angle DAB$
$\angle CAB=61+30.5$
$\angle CAB={91.5}^{\circ }$