5. If the cosines of the three interior angles of $\triangle A_{1} B_{1} C_{1}$ are equal to the sines of the three interior angles of $\triangle A_{2} B_{2} C_{2}$, then
A. $\triangle A_{1} B_{1} C_{1}$ is an acute triangle, and $\triangle A_{2} B_{2} C_{2}$ is also an acute triangle
B. $\triangle A_{1} B_{1} C_{1}$ is an obtuse triangle, and $\triangle A_{2} B_{2} C_{2}$ is also an obtuse triangle
C. $\triangle A_{1} B_{1} C_{1}$ is an acute triangle, and $\triangle A_{2} B_{2} C_{2}$ is an obtuse triangle
D. $\triangle A_{1} B_{1} C_{1}$ is an obtuse triangle, and $\triangle A_{2} B_{2} C_{2}$ is an acute triangle
$$
\begin{array}{l}
\text { Given }\left\{\begin{array} { l } 
{ \cos A _ { 1 } = \sin A _ { 2 } = \cos ( 90 ^ { \circ } - A _ { 2 } ) , } \\
{ \cos B _ { 1 } = \sin B _ { 2 } = \cos ( 90 ^ { \circ } - B _ { 2 } ) , } \\
{ \cos C _ { 1 } = \sin C _ { 2 } = \cos ( C _ { 2 } - 90 ^ { \circ } ) , }
\end{array} \Rightarrow \left\{\begin{array}{l}
A_{1}=90^{\circ}-A_{2}, \\
B_{1}=90^{\circ}-B_{2}, \\
C_{1}=C_{2}-90^{\circ}
\end{array}\right.\right. \\
\Rightarrow 90^{\circ}-A_{2}-B_{2}+C_{2}=180^{\circ} \Rightarrow-A_{2}-B_{2}+C_{2}=90^{\circ} \Rightarrow-A_{2}-B_{2}-C_{2}+2 C_{2}=90^{\circ} \\
\Rightarrow 2 C_{2}=270^{\circ} \Rightarrow C_{2}=135^{\circ}, \text { so choose } C .
\end{array}
$$