How to find $\angle$ b? How to find  $\angle$ b ?
The vertices of the triangle are on the foci of the ellipse and on the ellipse. 
$\angle$ a, the major axis and eccentricity are known.


Let $\alpha, \beta$ be the angles, $2a>0$ the length of the major axis, $P = (-ae, 0)$, $Q = (ae, 0)$ the foci, and $R = (x,y)$ the point on the ellipse. I assume for simplicity that $-ae < x < ae$ and $y >0$, since the calculations are analogous in other cases.
Note that $x,y$ are such that
\begin{align}
\frac{x^2}{a^2} + \frac{y^2}{(1-e^2)a^2} = 1
\end{align}
and
\begin{align*}
y = \tan \alpha (x+ae).
\end{align*}
We have
\begin{align*}
\tan \beta = \frac{y}{ae - x} = \frac{ae + x}{ae - x}\tan \alpha.
\end{align*}
Let $t = (ae+x)/(ae-x)$, so that $x = ae(t-1)/(t+1)$ and $y = ae\tan \alpha (2t)/(t+1)$. Then the ellipse equation for $t$ is
\begin{align*}
e^2\frac{(t-1)^2}{(t+1)^2} + 4\tan^2 \alpha \frac{e^2}{1-e^2}\frac{t^2}{(1+t)^2} = 1.
\end{align*}
If my calculations are correct, from this one gets
\begin{align*}
t = \frac{1-e^2}{4e^2\tan^2 \alpha - (1-e^2)^2}\left ( 1 + e^2 + \frac{2e}{\cos \alpha} \right ).
\end{align*}
This times $\tan \alpha$ gives $\tan \beta$. Of course the value of $\alpha$ such that $2e\tan \alpha = 1-e^2$ is the one for which $x = ae$, since $t \to \infty$ gives $x \to ae$. Note also that for $\tan^2 \alpha = (1-e^2)/e^2$, i.e. when $R = (0, b)$, the expression simplifies to $t = 1$, as expected.
Unfortunately I cannot write the solution in a better form than this... Hope it can help anyway! 
