Answer:
[tex]\angle 1=22, \angle\ 2=68, \angle 3=68, \angle 4=90[/tex]
Step-by-step explanation:
[tex]We\ are\ given:\\The\ quadrilateral\ is\ a\ rhombus.\\We\ know\ that,\\A\ rhombus\ is\ a\ parallelogram\ with\ all\ sides\ equal.\\As\ it\ is\ a\ parallelogram,\ the\ opposite\ sides\ are\ parallel\ too.\\Hence,\\\angle 3\ and\ \angle68\ form\ alternate\ interior\ angles\ to\ one\ pair\ of\\ opposite\ parallel\ sides\ in\ the\ rhombus,\ and\ are\ equal\ in\ measure.\\Hence,\\\angle 3=68\\[/tex]
[tex]As\ we\ know\ that\ the\ diagonals\ of\ a\ rhombus\ are\ perpendicular\\ bisectors\ to\ each\ other:\\\angle 4\ forms\ a\ 90.\\Hence,\\\angle 4=90[/tex]
[tex]Now,\ lets\ consider\ the\ isosceles\ triangle\ created\ by\ two\ equal\ sides\ of\\ the\ rhombus\ with\ one\ of\ its\ base\ angles\ \angle 2.\\Now,\\We\ know\ that,\\Base\ angles\ opposite\ to\ equal\ sides\ are\ equal\ too.\\Hence,\\As\ the\ two\ sides\ are\ equal,\\Base\ angles\ \angle 68\ and\ \angle 2\ are\ equal\ too.\\Hence,\\\angle 2=68[/tex]
[tex]Now,\ lets\ consider\ the\ triangle\ made\ up\ of\ \angle 1\ and\ \angle 2:\\We\ know\ that,\\\angle 2=68 [Proven]\\The\ third\ angle\ is\ 90\ as\ the\ diagonals\ of\ a\ rhombus\ are\ perpendicular\\ bisectors.\\Hence,\\\angle 1+ \angle 2 +90=180 [Angle\ Sum\ Property\ Of\ A\ Triangle]\\Hence,\\\angle 1+ \angle 2=90\\\angle 1+68=90\\\angle 1=90-68\\\angle 1=22[/tex]
[tex]By\ putting\ it\ all\ together\ we\ have:\\\angle 1=22, \angle\ 2=68, \angle 3=68, \angle 4=90[/tex]
