How would I go about solving this equation for $ \alpha $?
$$ \tan (\alpha + \Theta _1) - \tan (\alpha + \Theta _0) = \tan (\alpha + \Theta _0) - \tan (\alpha + \Theta _2) $$
Is there a way to calculate $\alpha$ directly or do I have to iterate?
$\alpha$ is always between $0$ and $\pi / 2$
Here is an abstract version of what those angles are
In case someone tumbles upon this: Check the answer of Barry Cipra. An alternative is to just brute force it the way I did and use the angle addition theorem like suggested in the comments. My solution was $$ \tan(\Theta_0) = c_0 $$ $$ \tan(\Theta_1) = c_1 $$ $$ \tan(\Theta_2) = c_2 $$ $$ \alpha = \arctan\left(2c_0 -c_1 - c_2 \over c_0c_2+c_1c_0 - 2c_1c_2\right) $$
which gives back the same result.
Let $x=\tan(\alpha+\Theta_0)$ and $T_i=\tan(\Theta_i-\Theta_0)$ for $i=1$ and $2$. Using the angle addition formula for the tangent function, we can rewrite the equation as
$${x+T_1\over1-xT_1}+{x+T_2\over1-xT_2}=2x$$
Clearing out denominators gives
$$(x+T_1)(1-xT_2)+(x+T_2)(1-xT_1)=2x(1-xT_1)(1-xT_2)$$
which expands to
$$(T_1+(1-T_1T_2)x-T_2x^2)+(T_2+(1-T_1T_2)x-T_1x^2)=2T_1T_2x^3-2(T_1+T_2)x^2+2x$$
and reduces to the cubic
$$2T_1T_2x^3-(T_1+T_2)x^2+2T_1T_2x-(T_1+T_2)=0$$
which factors as
$$(2T_1T_2x-T_1-T_2)(x^2+1)=0$$
The sole real root is
$$x={T_1+T_2\over2T_1T_2}$$
so up to an adjustment by $\pi$, we have
$$\alpha=\arctan\left(T_1+T_2\over2T_1T_2\right)-\Theta_0$$
Note, without some additional assumptions on the $\Theta_i$'s, we can only guarantee $|\alpha|\le\pi/2$, not $0\lt\alpha\lt\pi/2$
Added later (at OP's request): It takes a couple of steps to turn the equation
$$\tan(\alpha+\Theta_1)-\tan(\alpha+\Theta_0)=\tan(\alpha+\Theta_0)-\tan(\alpha+\Theta_2)$$
into the equation in $x$ and $T_i$s at the top of the answer. The first step is to move the subtracted tan's from one side to the other to get
$$\tan(\alpha+\Theta_1)+\tan(\alpha+\Theta_2)=2\tan(\alpha+\Theta_0)$$
The right hand side is now simply $2x$ when we let $x=\tan(\alpha+\Theta_0)$. The next trick is to note that $(\alpha+\Theta_i)=(\alpha+\Theta_0)+(\Theta_i-\Theta_0)$, and therefore
$$\tan(\alpha+\Theta_i)={\tan(\alpha+\Theta_0)+\tan(\Theta_i-\Theta_0)\over1-\tan(\alpha+\Theta_0)\tan(\Theta_i-\Theta_0)}={x+T_i\over1-xT_i}$$
when we let $T_i=\tan(\Theta_i-\Theta_0)$.
Remark: My initial impulse was to use the slightly simpler-looking substitutions $x=\tan\alpha$ and $T_i=\tan\Theta_i$, which turns the equation $\tan(\alpha+\Theta_1)+\tan(\alpha+\Theta_2)=2\tan(\alpha+\Theta_0)$ into
$${x+T_1\over1-xT_1}+{x+T_2\over1-xT_2}=2{x+T_0\over1-xT_0}$$
This also leads to a cubic after clearing out denominators, but there are three denominators to clear out, which means there's a lot of algebra to do. Being both lazy and error-prone, I discarded all that scratch work and started over with the substitution $x=\tan(\alpha-\Theta_0)$, which simplified the algebra and led to a nice result.