Question

$\text{The given figure shows a rhombus ABCD in which angle}BCD={180}^{\circ }.\text{Find angle s x and y.}$

Answer 1

$\text{In rhombus ABCD, diagaonals AC and BD bisect each other at}{90}^{\circ }\angle BCD={80}^{\circ }\because \text{Diagonals bisect the opposite angles also}\angle BCD=\angle BAD\left(\text{Opposite angles of rhombus}\right)\therefore \angle BAD={80}^{\circ }and\angle ABC=\angle ADC={180}^{\circ }-{80}^{\circ }={100}^{\circ }\therefore \text{Diagonals bisect opposite angles}\therefore \angle OCBor\angle PCB=\frac{{80}^{\circ }}{2}={40}^{\circ }In\Delta PCM,Ext.CPD=\angle OCB+\angle PMC{100}^{\circ }={40}^{\circ }+x\Rightarrow x={110}^{\circ }-{40}^{\circ }={70}^{\circ }and\angle ADO=\frac{1}{2}\angle ADC=\frac{1}{2}\times {100}^{\circ }={50}^{\circ }Hencex={70}^{\circ }andy={50}^{\circ }$