Answer:
[tex]\large\boxed{23^o20'48''=\left(23\dfrac{26}{75}\right)^o}[/tex]
Step-by-step explanation:
[tex]\text{We know:}\\\\1^o=60'\to1'=\left(\dfrac{1}{60}\right)^o\\\\1'=60''\to1^o=(60)(60'')=3600''\to1''=\left(\dfrac{1}{3600}\right)^o\\\\23^o20'48''\\\\20'=\left(\dfrac{20}{60}\right)^o=\left(\dfrac{1}{3}\right)^o\\\\48''=\left(\dfrac{48}{3600}\right)^o=\left(\dfrac{1}{75}\right)^o\\\\\text{Therefore}\\\\23^o20'48''=23^o+20'+48''=23^o+\left(\dfrac{1}{3}\right)^o+\left(\dfrac{1}{75}\right)^o\\\\=\left(23+\dfrac{1\cdot25}{3\cdot75}+\dfrac{1}{75}\right)^o=\left(23+\dfrac{25}{75}+\dfrac{1}{75}\right)^o=\left(23\dfrac{26}{75}\right)^o[/tex]
