Question

$ABCD$ is a rhombus. If $\angle BAC={38}^o$, find
$\angle ADC$

Answer 1

$\angle DAC=\angle BAC$ (diagonal bisect the angle)

$\begin{array}{c}\angle DAC={30}^{\circ }\\ \angle BAC=\angle ACD\end{array}$ (alternate angle)

$\begin{array}{c}\angle ACD={38}^{\circ }\\ In\Delta ADC\\ \angle DAC+\angle ACD+\angle ADC={180}^{\circ }\\ \Rightarrow {38}^{\circ }+{38}^{\circ }+\angle ADC={180}^{\circ }\\ \Rightarrow \angle ADC=180-76={104}^{\circ }\end{array}$