Question

In a rhombus $ABCD,\angle BAC={38}^{\circ }$. Find (i) $\angle ACD$, (ii) $\angle DAC$ and (iii) $\angle ADC$

Answer 1

We know that $\angle BAC={38}^{\circ }$.

But in a rhombus $ABCD$, since $ABC$ is an isosceles triangle, we see that $\angle BAC=\angle ACB={38}^{\circ }$.

Moreover $\angle DAC={38}^{\circ }$, since the diagonal $AC$ bisect $\angle A$.

Since $ADC$ is also an isosceles triangle, we get $\angle ACD=\angle DAC={38}^{\circ }$ ....... $\text{(i)}$

Finally,

$\angle ADC={180}^{\circ }-(\angle DAC+\angle DCA)$

$={180}^{\circ }-({38}^{\circ }+{38}^{\circ })$

$={180}^{\circ }-{76}^{\circ }$

$={104}^{\circ }$.

Thus, $\angle ADC={104}^o$