Đáp án
a) Ta có:
\(\eqalign{
& {21^0}30' = {{21,5\pi } \over {180}} \approx 0,375\,\,(rad) \cr
& {75^0}54' = {{75,9\pi } \over {180}} \approx 1,325\,\,(rad) \cr} \)
b) Ta có:
\(\eqalign{
& 2,5\,rad\,\, = \,{({{2,5.180} \over \pi })^0} \approx {143^0}14' \cr
& {2 \over \pi }rad\, = ({{{2 \over \pi }180} \over \pi }) = {({{360} \over \pi })^0} \approx {36^0}29' \cr} \)