Question

$\text{ABCD is a rhombus. if}\angle BAC={38}^{\circ },find:\left(i\right)\angle ACB\left(ii\right)\angle DAC\left(iii\right)\angle ADC.$

Answer 1

$\because \text{ABCD is Rhombus (Given)}\therefore AB=BC\therefore \angle BAC=\angle ACB\left(\angle \text{opp. to equal sides}\right)But\angle BAC={38}^{\circ }\left(Given\right)\therefore \angle ACB={38}^{\circ }In\Delta ABC,\angle ABC+BAC+\angle ACB={180}^{\circ }\angle ABC+{38}^{\circ }+{38}^{\circ }={180}^{\circ }\angle ABC={180}^{\circ }-{76}^{\circ }={104}^{\circ }But\angle ABC=\angle ADC\left(\text{oppl}angles\text{of hrombus}\right)\therefore \angle ADC={104}^{\circ }\because \angle DAC=\angle DCA(\because AD=CD)\therefore \angle DAC=\frac{1}{2}[{180}^{\circ }={104}^{\circ }]\angle DAC=\frac{1}{2}\times {76}^{\circ }={38}^{\circ }Hence\left(i\right)\angle ACB={38}^{\circ }\left(ii\right)\angle DAC={38}^{\circ }\left(iii\right)\angle ADC={104}^{\circ }$