Question

Por favor me ajudem, rápido!!

Answer 1

Olá, Júlia.

Se os ângulos pintados da mesma cor são suplementares, então temos que:

$\hat A + 72\º46'11''=180\º\Rightarrow\hat A = 179\º59'60''-72\º46'11''=\boxed{107\º13'49''}\\\\ \hat B + 152\º57''=180\º\Rightarrow\hat B = 179\º59'60''-152\º0'57''=\boxed{27\º59'3''}\\\\ \hat C + 103\º=180\º\Rightarrow\hat C = 180\º-103\º=\boxed{77\º}\\\\ \hat D + 128\º19'=180\º\Rightarrow\hat B = 179\º60'-128\º19'=\boxed{51\º41'}$

Answer 2

Se dois ângulos $\alpha$ e $\beta$ são suplementares, então
 
$\alpha+\beta=180^{\circ}$.

Com isso, temos que:

$\rhd$ $72^{\circ}46'11"+\hat{A}=180^{\circ}~\Rightarrow~\hat{A}=180^{\circ}-72^{\circ}46'11"$

Como $1^{\circ}=60'$, então, $180^{\circ}=179^{\circ}60'$.

Mas, $1'=60''$. Assim, $180^{\circ}=179^{\circ}59'60"$.

Com isso, $\hat{A}=179^{\circ}59'60"-72^{\circ}46'11"=107^{\circ}13'49"$.


$\rhd$ $\hat{B}+152^{\circ}57"=180^{\circ}~\Rightarrow~\hat{B}=180^{\circ}-152^{\circ}57"$.

Lembrando que, $180^{\circ}=179^{\circ}60'=179^{\circ}59'60"$, obtemos

$\hat{B}=179^{\circ}59'60"-152^{\circ}57"=27^{\circ}59'3"$.


$\rhd$ $103^{\circ}+\hat{C}=180^{\circ}~\Rightarrow~\hat{C}=180^{\circ}-103^{\circ}=77^{\circ}$.


$\rhd$  $128^{\circ}19'+\hat{D}=180^{\circ}~\Rightarrow~\hat{D}=180^{\circ}-128^{\circ}19'$.

Lembrando que, $180^{\circ}=179^{\circ}60'$, obtemos

$\hat{D}=179^{\circ}60'-128^{\circ}19'=51^{\circ}41'$.