Answer
$40^{\circ} 19^{'} 12^{"}$
Work Step by Step
RECALL:
$1^{\circ} = 60^{'} \text{ and } 1^{'} = 60^{"}$
\begin{equation}
\begin{split}
40.32^{\circ} & = 40^{\circ}+0.32^{\circ} \\ & = 40^{\circ}+0.32 \times(1^{\circ}) \\ & = 40^{\circ}+0.32 \times (60^{'}) \\ & = 40^{\circ}+19.2 ^{'} \\ & = 40^{\circ}+19^{'}+0.2{'} \\ & = 40^{\circ}+19^{'}+0.2 \times (1)^{'} \\ & = 40^{\circ}+19^{'}+0.2 \times (60)^{"} \\ & = 40^{\circ}+19^{'}+12^{"}
\end{split}
\end{equation}
Thus, $40.32^{\circ} = 40^{\circ} 19^{'} 12^{"}$
