How to Determine the Angle Between Normal and Full Acceleration of a Body?

A body of mass $m$ is thrown at an angle $alpha$ to the horizon with an initial velocity $v_o$. Determine the angle between the normal and total acceleration of the body, if the body is affected by the resistance force of the medium $F=-kv$.

I know this site is not really for homework, but the task is difficult and I’m not quite sure how to solve it, I reasoned like this:

It looks like we are being asked to find the angle at the moment of the throw, and not at any moment. Well .. anyway, we need to solve differential equations and find velocities. Knowing the speed, we immediately find the acceleration. Then we can rely on vectors. Let’s write acceleration as the sum of tangential and normal:
$w=wt+wn$

Further in the sense of tangent acceleration: $wt = (w, v) v / | v | ^ 2$

Then the normal acceleration is: $wn=w-(w,v)v/|v|^2$

The square of its modulus: $| wn | ^ 2 = (wn, wn) = | w | ^ 2 – (w, v) ^ 2 / | v | ^ 2$

And then we can write the dot product twice:

$(wn, w) = | w | ^ 2 – (w, v) ^ 2 / | v | ^ 2$

$(wn, w) = | w | | wn | cos(x)$

From here we express the cosine of the desired angle: $cos(x)=sqrt{(1-(w,v)^2/[|w||v|]^2)}$

Now I just need to substitute vectors of speed and acceleration and calculate? Am I reasoning correctly?