I haven't really found any quick, nice geometrical interpretation, so we need to do some calculations. It isn't too bad, though. For the sake of simplicity we will first assume that the ellipse is centered in $0$.
For an ellipse we get the following equation (since you are interested in angles, we will use polar coordinates):
$$X=\binom{h\cos \alpha}{k\sin\alpha}$$
At this stage you need to be aware that this gives you a parametrization of the curve, but does not tell you anything about what the angle $\alpha$ actually is, geometrically speaking. It turns out that this angle is not the possibly naturally assumed $\theta$ from your drawing. Luckily, it will turn out that we can get a fairly nice formula for that - if you are interested, you can find references to this under 'Eccentric anomaly', which bugged Johannes Kepler quite a lot, apparently.
So, first we will ignore this, use our parametrisation and get a nice result. Second, we will try to change our result to fit what you want.
First step:
Now, in order to get the tangent for a certain point, we derive this with respect to $\alpha$:
$$\frac{\partial X}{\partial\alpha}=\binom{-h\sin\alpha}{k\cos\alpha}$$
This gives us a directional vector of the tangent at the point $X(\alpha )$, or, to put it differently, of your line $c$.
Now, as $d$ is perpendicular to $c$, we get an equation for $d$ or the form
$$ d: \quad -h\sin\alpha x+k\cos\alpha y= z\qquad \Leftrightarrow \\
y=\frac{h\sin\alpha}{k\cos\alpha}x+\overline{z}=\frac{h}{k}\tan\alpha +\overline{z}$$
for some constants $z,\overline{z}$. And, seeing that the angle $\phi$ you want to find is nothing else than the tangent of the slope of $d$, we are finished and have $\tan\phi=\frac{h}{k}\tan\alpha$.
(Note that we assumed that $\cos\alpha\neq 0$, but that case is fairly simple - either you look at it geometrically, or just look at what happens to all those trignometric functions.)
Second Step:
Okay, so now we have expressed our result in terms of $\alpha$. That's, well, fairly nice but, as you pointed out, does not quite solve the provlem. Instead, we will have to take a look at how $\alpha$ is dependent on $\theta$. To do this, we need an equation.
If we look at the point of the ellipse determined by your angle $\theta$, it must be on the line $y=\tan\theta x$.
Similarly, for this point of the ellipse, we obviously have $\binom{x}{y}=\binom{h\cos\alpha}{k\sin\alpha}$.
Put together, this leaves us with the following system of equations:
$$y=x\tan\theta \\
x=h\cos\alpha \\
y=k\sin\alpha \\
$$
Now, we have two expressions for $y$, hence we have $x\tan\theta=k\sin\alpha$, and the second equation gives us $x=h\cos\alpha$, put together this leaves us with
$$h\cos\alpha\tan\theta=k\sin\alpha\qquad \Leftrightarrow \\
\tan\alpha=\frac{h}{k}\tan\theta$$
Notice that, this seems almost custom-tailored to what we want, because in the result of the first part, all we need is $\tan\alpha$, had we obtained some more obscure expression, well, then we might still had a bit more work to do.
Conclusion
Putting those two results together, one now obtains $\tan\phi=\frac{h}{k}\tan\alpha=\frac{h^2}{k^2}\tan\theta$.
Looking at this result, I actually don't find it as intuitively clear as I thought the wrong version to be. However, in a weird way, it does make kind of sense. That the angle scales with the value of $\frac{h}{k}$, certainly does, the square is not something I'd immediately think of. However, if you ever happen to do any kind of coordinate transformations, or path integrals, or anything like that, this is a fairly good example why one has all those weird transformation formulas.
Possibly something pointing in the direction of why the first approach cannot be right is if you simply take $\theta=\frac{pi}{4}$ (meaning $x=y$), and plug it all into your equation. If you substitute $h\cos\alpha = x$ and so on, you get the tangent vector $\binom{-h/k}{k/h}$, whereas - if $\alpha$ were $\theta$, we only get $\binom{-h}{k}$, which kind of does not work out since those two will, generally, not be parallel.
So, if you think of an ellipse as a kind of scaled circle, where we apply the weight $h$ to the cosine and $k$ to the sine, we "get" the actual weight $\frac{h^2}{k^2}$ for the tangent, and the fact that for $h=k$ we have $\phi=\theta$, which would be the relation in any ordinary circle, still holds. This is a bit similar to what happens if you look at your problem in a different coordinate system - that would also be a nice way to solve the question, involving nothing but a couple coordinate transformations, but possibly not feasible depending on your knowledge and what you need the problem for.
Now, if our ellipse is not centered at $0$ but aligned with the axes, nothing changes. Translation does not change angles. If we rotate our ellipse by any fixed angle $\gamma$, well, then you will need to subtract $\gamma$ from any angle you get as a result.
Best Answer
Steps hinted (justify/prove each step in the following):
The triangle $\;\Delta OCD\;$ is isosceles (why?), and thus $\;\angle OCD=\angle ODC\;$, and since $\;\angle DOC\;$ is the other angle , it must be that
$$111^\circ-x^\circ=\angle DOC=180^\circ-2\angle OCD$$
But $\;\angle OCD\;$ is an exterior one to $\;\Delta OCE\;$ and thus $\;\angle OCD=2x^\circ\;$ (why?), so we get
$$111^\circ-x^\circ=180-4x^\circ\implies3x^\circ=69^\circ\implies x=23^\circ$$