Đổi thành độ, phút
\(\eqalign{
& 15,{25^o} = 15{{{1^o}} \over 4} = {15^o}15′ = 915′ \cr
& 30,{5^o} = ................. = .................. = \cr
& 60,{75^o} = ................ = .................. = \cr
& 90,{2^o} = ................. = .................. = \cr
& 45,{15^o} = ................ = ................... = \cr} \)
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Giải
\(\eqalign{
& 15,{25^o} = 15{{{1^o}} \over 4} = {15^o}15′ = 915′ \cr
& 30,{5^o} = 30{{{1^o}} \over 2} = {30^o}30′ = 1830′ \cr
& 60,{75^o} = 60{{{3^o}} \over 4} = {60^o}45′ = 3645′ \cr
& 90,{2^o} = 90{{{1^o}} \over 5} = {90^o}12′ = 5412′ \cr
& 45,{15^o} = 45{{{3^o}} \over {20}} = {45^o}9′ = 2709′ \cr} \)