Denote $AB$ the straight line distance between points $A$ and $B$, $r$ the unknown radius of our circle, and $C$ the circle centre.
We thus have an isosceles triangle $ABC$, whose two sides are of length $r$ and the third side is length $AB$.
Let us draw two tangents to our circle - one at point $A$ and the other at point $B$. The tangent at point $B$ will be perpendicular to side $BC$, while the tangent at $A$ will be perpendicular to side $AC$. Thus the two angles $\angle ABC $ and $\angle BAC$ will be $(\frac{\pi}{2}-a)$
Using trigonometry, we obtain $AB=2r\sin a$ so that
$\large r=\frac{AB}{2\sin a}$
Now the angle $\angle BCA$ of our triangle is $2a$ (in radians), so that our arc length $x$ is
$\large x=2\pi r\frac{2a}{2\pi}=2ra=AB\frac{a}{\sin a}$
To find the angles at which the front and back wheels need to be set, let us assume that the wheels are spaced apart by distance $z$ (which is almost the length of our car). The wheels need to be set at tangents to the circle at the points at which they intersect the circle.
We set the back wheels to angle $a$ relative to line $AB$, assuming they are at point $A$. Now we need to find the angle at which to set the front wheels.
Let point $D$ be where the front wheels are - for simplicity we assume they lie on the arc.
We have another isosceles triangle $ACD$, with same angles $\angle CAD$ and $\angle CDA$.
Using trigonometry, we have
$\angle CAD$ = $\angle CDA = \arccos \frac{z}{2r}=\arccos \frac{z\sin a}{AB}$.
Then the angle $\angle DCA = \pi - 2\arccos \frac{z\sin a}{AB}$
If we draw a line from point $D$ that is parallel to $AC$, called $DE$ (where $E$ is left of $D$), the angle $\angle EDC = \angle DCA$, so that the angle the front wheels make relative to $DE$ is
$\large \frac{\pi}{2}-\angle EDC=2\arccos \frac{z\sin a}{AB}-\frac{\pi}{2}$
However it would be better to find the angle of the front wheel relative to a line parallel to $AB$. To do this, we identify a point $F$ to the left of $D$ so that line $FD$ is parallel to $AB$.
As $FD$ is parallel to $AB$ by construction, $\angle CDF+\angle CDA+\angle BAD=\pi$.
Now we have
$\angle BAD=\angle CAD - \angle CAB = \arccos \frac{z\sin a}{AB}-(\frac{\pi}{2}-a)=\arccos \frac{z\sin a}{AB}-\frac{\pi}{2}+a$
Thus
$\angle CDF = \pi - \angle CAD - \angle BAD= \pi-2\arccos \frac{z\sin a}{AB}+\frac{\pi}{2}-a$
This leads to the angle the front wheel makes to $DF$ (which is parallel to $AB$) being
$\large \frac{\pi}{2}-\angle CDF=a+2\arccos\frac{z\sin a}{AB}-\pi$
This result is quite interesting, because as $z$ tends to zero (the car shrinks),
$2\arccos\frac{z\sin a}{AB}\rightarrow \pi$
So that the angle the front wheels make will approach (in value) the angle the rear wheels make i.e. $a$.
If $z$ is non-zero but small, $2\arccos\frac{z\sin a}{AB}<\pi$, so that the front wheel angle will be slightly less than $a$, which is what our intuition would suggest.
Best Answer
If you want to stay within 10%, the following piecewise linear function satisfies $$.9\tan\theta \le y \le 1.1\tan\theta$$ for $0\le\theta\le60$ degrees:
$$y={\theta\over60}\text{ for }0\le\theta\le20$$ $$y={2\theta-15\over75}\text{ for }20\le\theta\le45$$ $$y={\theta-20\over25}\text{ for }45\le\theta\le60$$
It might help to rewrite them as
$$y={5\theta\over300}\text{ for }0\le\theta\le20$$ $$y={8\theta-60\over300}\text{ for }20\le\theta\le45$$ $$y={4\theta-80\over100}\text{ for }45\le\theta\le60$$
so that you really don't have to divide by anything other than $3$. The line segment approximations lie above $\tan\theta$ from $\theta\approx25$ to $\theta=45$ and below it elsewhere, so you should round down and up accordingly when doing the mental arithmetic.It's obviously possible to extend this for angles greater than $60$ degrees, but whether (or how far) you can do so with formulas that use only "simple" multiplications and divisions is unclear.
A word of explanation: What I tried to do here was take seriously the OP's request for estimates you can calculate in your head. The ability to do mental arithmetic, of course, varies from person to person, so I used myself as a gauge. As for where the formulas came from, my starting point was the observation that the conversion factor between degrees and radians, $180/\pi$, is approximately $60$, so the estimate $\tan\theta\approx\theta/60$ should be OK for a while. A little trial and error showed it's good up to $\theta=20$ degrees (since $.9\tan20\approx.328$). It was easy to see that connecting $(0,0)$ to $(20,1/3)$ and $(20,1/3)$ to $(45,1)$ with straight lines would stay within the prescribed bounds. Finally, noting that $.9\tan60\approx1.55$, I saw that the line connecting $(45,1)$ to $(60,1.6)$ would have a nice slope and stay within the prescribed bounds as well.