Question

In given figure $ABCD$ is a rhombus. If $\angle BAC={38}^o$, find:
$\angle DAC$

Answer 1

$\because ABCD$ is a rhombus.

$\therefore AB=BC$

$\Rightarrow \angle BAC=\angle ACB(\because \text{Angles opposite to equal sides are equal})$

$\because \angle BAC=38°\left(\text{Given}\right)$

$\therefore \angle ACB=38°$

In $△ABC$,

$\angle ABC+\angle BAC+\angle BCA=180°$

$\angle ABC+38°+38°=180°$

$\Rightarrow \angle ABC=180°-76°=104°$

$\because \angle ADC=\angle ABC(\because \text{Opposite angles of rhombus})$

$\therefore \angle ADC=104°$

$\angle DAC=\angle DCA(\because AD=CD)$

Now, in $△ACD$,

$\angle ADC+\angle ACD+\angle DAC=180°$

$\angle DAC+\angle DAC+104°=180°$

$2\angle DAC=180°-104°$

$\Rightarrow \angle DAC=\frac{76°}{2}=38°$