Question

matemática.

(regrinha: 1°=60'
1'=60")

(CALCULE)
45°20'25" - 30°30'30"=​

Answer 1

$45^\circ20'25''~-~30^\circ30'30''~=\\\\\\=~(45^\circ+20'+25'')~-~(30^\circ+30'+30'')\\\\\\Como~podemos~ver~25''~\acute{e}~menor~que~30'',~logo~25''~vai~precisar~''pegar\\emprestado''~de~20',~acompanhe:\\\\\\=~(45^\circ+19'+1'+25'')~-~(30^\circ+30'+30'')\\\\\\=~(45^\circ+19'+60''+25'')~-~(30^\circ+30'+30'')\\\\\\=~(45^\circ+19'+85'')~-~(30^\circ+30'+30'')$

$Agora~sim,~85''~\acute{e}~maior~que~30'',~logo~podemos~subtrair~os~dois\\numeros~normalmente:\\\\\\=~(45^\circ+19'+85''-30'')~-~(30^\circ+30')\\\\\\=~(45^\circ+19'+55'')~-~(30^\circ+30')\\\\\\Temos~agora~que~19'~\acute{e}~menor~que~30',~logo~vamos~''pegar~emprestado''\\novamente:$

$=~(44^\circ+1^\circ+19'+55'')~-~(30^\circ+30')\\\\\\=~(44^\circ+60'+19'+55'')~-~(30^\circ+30')\\\\\\=~(44^\circ+79'+55'')~-~(30^\circ+30')\\\\\\Efetuando~as~subtracoes:\\\\\\44^\circ-30^\circ+79'-30'+55''\\\\\\=~\boxed{14^\circ+49'+55''}~~~ou~~~\boxed{14^\circ49'55''}$