Triangle \[\triangle A'B'C'\] is the image of \[\triangle ABC\] under a rotation about the origin, \[(0,0)\]. \[\small{1}\] \[\small{2}\] \[\small{3}\] \[\small{4}\] \[\small{5}\] \[\small{6}\] \[\small{7}\] \[\small{\llap{-}2}\] \[\small{\llap{-}3}\] \[\small{\llap{-}4}\] \[\small{\llap{-}5}\] \[\small{\llap{-}6}\] \[\small{\llap{-}7}\] \[\small{1}\] \[\small{2}\] \[\small{3}\] \[\small{4}\] \[\small{5}\] \[\small{6}\] \[\small{7}\] \[\small{\llap{-}2}\] \[\small{\llap{-}3}\] \[\small{\llap{-}4}\] \[\small{\llap{-}5}\] \[\small{\llap{-}6}\] \[\small{\llap{-}7}\] \[y\] \[x\] \[A\] \[B\] \[C\] \[A'\] \[B'\] \[C'\] Determine the angle of rotation. Choose 1 answer: Choose 1 answer: (Choice A) \[-165^\circ\] A \[-165^\circ\] (Choice B) \[-135^\circ\] B \[-135^\circ\] (Choice C) \[135^\circ\] C \[135^\circ\] (Choice D) \[165^\circ\] D \[165^\circ\]
To find the angle of rotation, we can examine the position of points $A$, $A'$ and $B$, $B'$. We see that point $A'$ is the image of point $A$ after rotating $180^\circ$ counterclockwise about the origin. Similarly, point $B'$ is the image of $B$ after rotating $180^\circ$ counterclockwise about the origin. Therefore, the triangle formed by points $A'$, $B'$ and the origin is a reflection of triangle $ABC$ over the $x$-axis.
[asy]
unitsize(0.6 cm);
dot((7,0));
dot((0,0));
dot((-7,0));
dot((0,-3));
dot((0,-7));
draw((-8.5,0)--(8.5,0));
draw((0,-8.5)--(0,1));
draw((-7.7,-2.3)--(7.7,2.3),dashed);
label("$A$", (7,0), E);
label("$B$", (0,-3), S);
label("$C$", (0,0), NW);
label("$A'$", (-7,0), W);
label("$B'$", (0,-7), E);
[/asy]
To form $\triangle A'B'C'$, we need to rotate $\triangle A'B'C'$ another $180^\circ$ counterclockwise about the origin. Since a $180^\circ$ rotation is the same as a reflection, $\triangle A'B'C'$ is the reflection of triangle $ABC$ over the $x$-axis.
Drawing a vertical line through points $A$ and $C$, we find that the vertical distance between $A$ and the $x$-axis is equal to the vertical distance between $C$ and the $x$-axis. This means that triangle $ACB$ is an isosceles triangle with $AC=BC$.
[asy]
unitsize(0.6 cm);
dot((7,0));
dot((0,0));
dot((-7,0));
dot((0,-3));
dot((0,-7));
draw((-8.5,0)--(8.5,0));
draw((0,-8.5)--(0,1));
draw((-7.7,-2.3)--(7.7,2.3),dashed);
label("$A$", (7,0), E);
label("$B$", (0,-3), S);
label("$C$", (0,0), NW);
label("$A'$", (-7,0), W);
label("$B'$", (0,-7), SE);
draw((-7,0)--(7,0));
draw((7,0)--(0,-7),dashed);
draw((0,0)--(0,-7),dashed);
[/asy]
After reflecting triangle $ABC$ over the $x$-axis to form $\triangle A'B'C'$, the lengths of the corresponding sides are preserved. Therefore, triangle $A'B'C'$ is also isosceles with $A'C'=B'C'$.
[asy]
unitsize(0.6 cm);
dot((7,0));
dot((0,0));
dot((-7,0));
dot((0,-3));
dot((0,-7));
draw((-8.5,0)--(8.5,0));
draw((0,-8.5)--(0,1));
draw((-7.7,-2.3)--(7.7,2.3),dashed);
draw((-7,0)--(0,7),red);
draw((0,-3)--(0,0),dashed);
draw((7,0)--(0,-7),dashed);
label("$A$", (7,0), E);
label("$B$", (0,-3), S);
label("$C$", (0,0), NW);
label("$A'$", (-7,0), W);
label("$B'$", (0,-7), SE);
label("$C'$", (0,7), N);
[/asy]
Since $AC=BC$ and $A'C'=B'C'$, we can conclude that triangle $BAC'$ is congruent to triangle $B'AC'$. Therefore, $\angle BAC'=\angle B'AC'$.
Finally, we see that $\angle BAC' + \angle A'AC' + \angle CAB = 180^\circ$ and $\angle B'AC' + \angle A'AC' + \angle C'AB' = 180^\circ$, so
\begin{align*}
\angle BAC' &= 180^\circ - \angle A'AC' - \angle CAB \\
&= 180^\circ - \angle A'AC' - \angle C'AB' \\
&= \angle B'AC'.
\end{align*}Since $\angle BAC' = \angle B'AC'$, we can conclude that $\angle CAB = \angle C'AB'$.
Therefore, $\triangle ABC$ is related to $\triangle A'B'C'$ by a $180^\circ$ rotation about the origin. This rotation can also be described as a reflection over the $x$-axis followed by a reflection over the $y$-axis. Such a rotation is equivalent to a $180^\circ$ rotation about the origin.
In other words, $\triangle A'B'C'$ is formed by rotating $\triangle ABC$ by $\boxed{180^\circ}$.