Triangle RST is drawn on the coordinate plane with vertices at (5,−6) , (5,0) , and (7,2) . The triangle is then translated 2 units to the right and 4 units down to create triangle R′S′T′ . What is the sum of the angles of triangle R′S′T′ ?
We see that triangle $RST$ is an isosceles right triangle with a right angle at $S$.
[asy]
unitsize(0.5 cm);
pair R, S, T, R', S', T';
R = (5,-6);
S = (5,0);
T = (7,2);
R' = shift((2,-4))*(R);
S' = shift((2,-4))*(S);
T' = shift((2,-4))*(T);
draw(R--S--T--cycle);
draw(R'--S'--T'--cycle,red);
draw((-10,0)--(10,0));
draw((0,-10)--(0,10));
label("$R$", R, SW);
label("$S$", S, N);
label("$T$", T, NE);
label("$R'$", R', W);
label("$S'$", S', S);
label("$T'$", T', E);
label("$(5,-6)$", R, NW);
label("$(5,0)$", S, N);
label("$(7,2)$", T, NE);
dot("$2$", midpoints(S--S'), N);
dot("$4$", midpoints(S--S'), E);
[/asy]
Translating $RST$ does not change the angles, so triangle $R'S'T'$ is also an isosceles right triangle. The sum of the angles of an isosceles triangle is $180 - 90 = 90$. Thus, the sum of the angles of triangle $R'S'T'$ is $\boxed{90}$.