### Instruction:
In the figure, $PA$ is tangent to semicircle $SAR$, $PB$ is tangent to semicircle $RBT$, and $SRT$ is a straight line.  If arc $AS$ is $58^\circ$ and arc $BT$ is $37^\circ$, then find $\angle APB$, in degrees.

[asy]
import graph;

unitsize(1.5 cm);

pair A, B, P, R, S, T;
pair[] O;
real[] r;

r[1] = 1;
r[2] = 0.8;

S = (-2*r[1],0);
O[1] = (-r[1],0);
R = (0,0);
O[2] = (r[2],0);
T = (2*r[2],0);
A = O[1] + dir(180 - 58)*r[1];
B = O[2] + dir(37)*r[2];
P = extension(A, A + rotate(90)*(A - O[1]), B, B + rotate(90)*(B - O[2]));

draw(S--T);
draw(arc(O[1],r[1],0,180));
draw(arc(O[2],r[2],0,180));
draw(A--P--B);

label("$A$", A, NW);
label("$B$", B, NE);
label("$P$", P, N);
label("$R$", R, dir(270));
label("$S$", S, SW);
label("$T$", T, SE);
[/asy]

### Response:

Let $O_1$ be the center of semicircle $SAR$, and let $O_2$ be the center of semicircle $RBT$.

[asy]
import graph;

unitsize(1.5 cm);

pair A, B, P, R, S, T;
pair[] O;
real[] r;

r[1] = 1;
r[2] = 0.8;

S = (-2*r[1],0);
O[1] = (-r[1],0);
R = (0,0);
O[2] = (r[2],0);
T = (2*r[2],0);
A = O[1] + dir(180 - 58)*r[1];
B = O[2] + dir(37)*r[2];
P = extension(A, A + rotate(90)*(A - O[1]), B, B + rotate(90)*(B - O[2]));

draw(S--T);
draw(arc(O[1],r[1],0,180));
draw(arc(O[2],r[2],0,180));
draw(A--P--B);
draw(A--O[1]);
draw(B--O[2]);

label("$A$", A, NW);
label("$B$", B, NE);
label("$O_1$", O[1], dir(270));
label("$O_2$", O[2], dir(270));
label("$P$", P, N);
label("$R$", R, dir(270));
label("$S$", S, SW);
label("$T$", T, SE);
[/asy]

Since $\angle AO_1 S = 58^\circ$, $\angle AO_1 R = 180^\circ - 58^\circ = 122^\circ$.  Since $\angle BO_2 T = 37^\circ$, $\angle BO_2 R = 180^\circ - 37^\circ = 143^\circ$.

The angles of pentagon $AO_1 O_2 BP$ add up to $540^\circ$, so \begin{align*}
\angle APB &= 540^\circ - \angle PAO_1 - \angle AO_1 R - \angle BO_2 R - \angle PBO_2 \\
&= 540^\circ - 90^\circ - 122^\circ - 143^\circ - 90^\circ \\
&= 95^\circ.
\end{align*}