In the diagram x + y = 220o . Find the value of n
20o
40o
60o
80o
$\begin{align} & y+\angle CBD={{180}^{\circ }}\text{ }\!\!\{\!\!\text{ sum of }\angle \text{s on a straight line }\!\!\}\!\!\text{ } \\ & \angle CBD={{180}^{\circ }}-y \\ & x+\angle BDC={{180}^{\circ }}\text{ }\!\!\{\!\!\text{ sum of }\angle \text{s on a straight line }\!\!\}\!\!\text{ } \\ & \angle BDC={{180}^{\circ }}-x \\ & \angle CBD+\angle BDC+\angle BCD={{180}^{\circ }} \\ & {{(180-y)}^{\circ }}+{{(180-x)}^{\circ }}+n={{180}^{\circ }}\text{ }\!\!\{\!\!\text{ sum of }\angle \text{s in a }\vartriangle \text{ }\!\!\}\!\!\text{ } \\ & \text{36}{{\text{0}}^{\circ }}-x-y+n={{180}^{\circ }} \\ & \text{36}{{\text{0}}^{\circ }}-(x+y)+n={{180}^{\circ }} \\ & {{360}^{\circ }}-{{220}^{\circ }}+n={{180}^{\circ }} \\ & {{140}^{\circ }}+n={{180}^{\circ }} \\ & n={{40}^{\circ }} \\\end{align}$
