When you type
asind(1.0019)
you are asking for the angle (in degrees) whose sine is equal to 1.0019. But no real-valued angle exists whose sine has that value.
The inverse sine for complex numbers can be defined as
f_inverse_sine = @(z)(-i*log(i*z+sqrt(1-z.^2)))
(I've switched over to radians here, but you could multiply the result by 180/pi to get degrees.) MATLAB has simply done the right thing for you in the complex domain.
But, since you are working with triangles, I am guessing you have just made a mistake that ended up with your taking the arcsin of something greater than 1, when you really shouldn't be.
