Question

Find the supplementary of ${37}^0{28}^{\prime }$ and ${52}^0{33}^{\prime }$.

Answer 1

$Foranangletobesupplementary,thesumoftwoanglesshouldbe{90}^{\circ }.$

$Supplementaryof{37}^{\circ }{28}^{{}^{\prime }}=90-{37}^{\circ }{28}^{{}^{\prime }}={52}^{\circ }{32}^{{}^{\prime }}$

$Supplementaryof{52}^{\circ }{33}^{{}^{\prime }}=90-{52}^{\circ }{33}^{{}^{\prime }}={37}^{\circ }{27}^{{}^{\prime }}$