Answer
$\angle1\ is112^o$
$\angle2\ is\ 68^o$
$\angle3\ is\ 68^o$
$\angle4\ is\ 112^o$
$\angle5\ is\ 112^o$
$\angle6\ is\ 68^o$
Work Step by Step
$\angle1\ is\ supplementary\ to\ \angle2 ; 180-68=112$
$\angle2\ is\ 68^o\ because\ it\ is\ a\ corresponding\ angle\ to\ \angle3$
$\angle3\ is\ 68^o$
$\angle4\ is\ supplementary\ to\ \angle3 ; 180-68=112$
$\angle5\ is\ supplementary\ to\ \angle6 ; 180-68=112$
$\angle6\ is\ 68^o$
