Hej,

I've modified your sketch a little bit and here is what I've calculated:

1. The straight line representing the place where your head could be has the equation $\displaystyle y = x \cdot \tan(22^\circ) - 3 \cdot \tan(22^\circ) +1$

$\displaystyle z = \alpha+\beta$

2. $\displaystyle \tan(\alpha)=\frac{10-y}{x}$ and $\displaystyle \tan(\beta) = \frac{y-2}x$

3. From $\displaystyle \tan(z)=\tan(\alpha + \beta)=\frac{\tan(\alpha) + \tan(\beta)}{1-\tan(\alpha) \cdot \tan(\beta)}$ I get:

$\displaystyle \tan(z)=\frac{\frac{10-y}{x} + \frac{y-2}x}{1 - \frac{10-y}{x} \cdot \frac{y-2}x}$

4. Since z has it's maximum if $\displaystyle \tan(z)$ is at it's maximum I solved for x: $\displaystyle \frac{d(\tan(z))}{dx} = 0$

I've got $\displaystyle x \approx 4.406798787$. That means your head is at (4.406798787, 1.568383604)

5. ... but honestly I would refuse to take this seat