\(\begin{array}{l}a)\;\;{18^0} = \frac{{18.\pi }}{{{{180}^0}}}\left( {rad} \right) = \frac{\pi }{{10}}\;\;\left( {rad} \right).\\b)\;\;{57^0}30' = 57,{5^0} = \frac{{57,5\pi }}{{{{180}^0}}}\left( {rad} \right) \\= \frac{{23}}{{72}}\left( {rad} \right).\\c) - {25^0} = \frac{{ - 25\pi }}{{{{180}^0}}}\left( {rad} \right) = - \frac{{5\pi }}{{36}}\left( {rad} \right).\\d) - {125^0}45' = - 125,{75^0} \\= - \frac{{125,75\pi }}{{{{180}^0}}}\left( {rad} \right) = - \frac{{503\pi }}{{720}}\left( {rad} \right).\end{array}\)
