You could try a substitution. For example, suppose $u = cos(\theta)$; then $\frac{sin(\theta)}{cos(\theta)} = \frac{\sqrt{1-u^2}}{u}$ and $\frac{1}{cos^2(\theta)} = \frac{1}{u^2}$. Your whole equation becomes the following:$$0=h+d\frac{\sqrt{1-u^2}}{u}+0.5a\frac{d^2}{u^2v^2}$$
Not what I'd call "pretty" by any stretch, but at least it's explicitly in only one unknown now. Clearing the denominators of $u$'s and solving for it yields:$$0=u^2h+du\sqrt{1-u^2}+\frac{0.5d^2a}{v^2}$$
$$-du\sqrt{1-u^2}=u^2h+\frac{0.5d^2a}{v^2}$$
$$u^2d^2(1-u^2)=(u^2h+\frac{0.5d^2a}{v^2})^2$$
$$-u^4d^2+u^2d^2=u^4h^2+\frac{u^2d^2ah}{v^2}+\frac{d^4a^2}{4v^4}$$
Let $z=u^2$. Rearranging to get everything on one side and making the substitution gets you the following:
$$(h^2+d^2)z^2+\frac{d^2(ah-v^2)}{v^2}z+\frac{d^4a^2}{4v^4}=0$$
$$z=\frac{\frac{d^2(v^2-ah)}{v^2}\pm\sqrt{\frac{d^4((ah-v^2)^2-a^2)}{v^4}}}{2(h^2+d^2)}$$
$$z=\frac{d^2((v^2-ah)\pm\sqrt{(ah-v^2)^2-a^2})}{2v^2(h^2+d^2)}$$
That's about as much as I have time for right now, I'm afraid, but you should get the gist by now. Presuming that the two values for $z$ are real, you'll be able to take the square root of the positive one to recover the values for $u$, and then take the arccos of those. Throw out whatever solutions don't correspond to the reality of the problem. As for finding $t$, all you need to do is compute $d/u$ for the values of $u$ you find; again, ignore any nonsensical results (like $t$ being negative for example).
Of course, all of this will only work if $(ah-v^2)^2-a^2 \ge 0$, which means that $(ah-v^2)^2 \ge a^2 \Rightarrow ah-v^2 \ge a \Rightarrow a(h-1)\ge v^2 \Rightarrow a\ge v^2/(h-1)$.
Addendum: taken to its ultimate conclusion, the values of $\theta$ can be computed with the given information using the following formula: $$\theta = \arccos(\pm\sqrt{\frac{d^2((v^2-ah)\pm\sqrt{(ah-v^2)^2-a^2})}{2v^2(h^2+d^2)}})$$
While the values of $t$ are given by:
$$t = \pm\frac{d}{\sqrt{\frac{d^2((v^2-ah)\pm\sqrt{(ah-v^2)^2-a^2})}{2v^2(h^2+d^2)}}}$$
I leave it to you to separate the valid cases from the invalid ones.