##-60°##
A right-angled triangle with hypotenuse ##1## and one side ##sqrt(3)/2## has the third side of length ##1/4## by Pythagoras. Draw a unit circle centred on the origin ##O## and sketch a regular hexagon in it, with vertices ##(1,0)##, ##(1/2,sqrt(3)/2)##, ##(-1/2, sqrt(3)/2)##, ##(-1,0)##, ##(-1/2,-sqrt(3)/2)##, ##(1/2,-sqrt(3)/2)##. Let ##P## the last vertex in this list and let ##N## be the foot of the perpendicular from ##P## to the ##x##-axis. Then, disregarding signs, angle ##hat{PON}=sin^-1(sqrt(3)/2)##. But clearly ##hat{PON}##=360°/6 by symmetry.
Applying the definition of “principal value” of the inverse sin function (which requires the angle to be within the inclusive range ##+-90°##) you get ##-60°##.
There are many other equivalent ways of visualizing this result.
