Edit
Original post at the end. This is uglier, but cleaner and more robust:
tmax = Pi/3;
rmax = 1.5;
u = PolarPlot[t (*Your function Here*), {t, 0, Pi},
PolarAxes -> {True, True}, PolarTicks -> {"Degrees", Automatic},
PolarGridLines -> True, PlotRange -> rmax,
RegionFunction -> Function[{x, y, t, r}, t < tmax],
PolarAxesOrigin -> {0, rmax}, PolarAxes -> 0];
Show[Quiet@
Replace[u,
{Circle[{0, 0}, x_, {0, 2 Pi}] -> Circle[{0, 0}, x, {0, tmax}],
Line[{x_, y_}] /; tmax < ArcTan[y[[1]], y[[2]]] || 0 > ArcTan[y[[1]], y[[2]]] :> {},
Line[{Scaled[x1_, y1_], Scaled[x2_, y2_]}] /;
tmax < ArcTan[y2[[1]] + x2[[1]], y2[[2]] + x2[[2]]] ||
0 > ArcTan[y2[[1]] + x2[[1]], y2[[2]] + x2[[2]]] :> {},
{{a_ (Sin | Cos)[y_], b_ (Sin | Cos)[y_]},
Scaled[{s_, t_}, {c_ (Sin | Cos)[y_], d_ (Sin | Cos)[y_]}]} /;
ArcTan[s, t] > tmax || ArcTan[s, t] < 0 -> {},
Text[Style[TraditionalForm[Times[x_, Degree]], List[]], __] /; x > tmax (180/Pi) :> {}},
Infinity],
PlotRange -> {{0, rmax}, {0, rmax Sin[tmax]}}]

Original post
I know this is no beauty, but just an idea:
rmax = 1.5;
Show[
PolarPlot[1.3 t (*Your function Here*), {t, 0, Pi},
PolarAxes -> {True, False},
PolarTicks -> {"Degrees", Automatic}, PolarGridLines -> True,
PlotRange -> rmax,
RegionFunction -> Function[{x, y, t, r}, t < Pi/3],
PolarAxesOrigin -> {0, rmax}],
Graphics[{White, Disk[{0, 0}, rmax 2, {0 - 1/15, -5/3 Pi + 1/20}]}],
PolarPlot[rmax, {t, 0, Pi/3}, PolarAxes -> {False, True},
PolarTicks -> {None, Automatic}, PolarGridLines -> False,
PlotRange -> rmax, PolarAxesOrigin -> {0, rmax},
PolarAxes -> 0]
,PlotRange -> {{0, rmax}, {0, rmax Sin[Pi/3]}}]

The
{0 - 1/15, -5/3 Pi + 1/20}
needs some elaboration, for sure.