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.

 A: Let the major axis be of length = $2a$, 
and the minor be of $2b$, then we know that 
$$b^2=a^2(1-e^2)$$
Foci are given by $P(ae,0)$,$Q(-ae,0)$
Let the point not on the vertex be $R(a\cos \theta,b\sin \theta)$
And I take angles RQP and RPQ as $\alpha$ and $\beta$ to avoid confusion.
Now we know that $$PQ=RQ \cos \alpha+RP \cos \beta$$
RQ and RP can be evaluated using the distance formula, 
We get $$RP=|a-ae\cos \theta| , RQ=|a+ae\cos \theta|$$
So now you say you know angle $a$ (i.e $\alpha$), also note that $a\cos \theta$
represents the x coordinate of the point R (not on the Foci) now we have all three sides of triangle PQR, applying the sine law in triangle PQR, we get
$$\frac{\sin \alpha}{RP}=\frac{\sin \beta}{RQ}$$
$$\implies \frac{\sin \alpha}{|a-ae\cos \theta|}=\frac{\sin \beta}{|a+ae\cos \theta|}$$
Now point R is also unknown to us, we can find it by considering the intersection of line QR with our ellipse,
i.e R will be point of intersection of:
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$
and
$$y=\tan\alpha (x+ae)$$
Solving this, you would get the coordinates of R.
As we only need the x coordinate, we can put that as  $a\cos \theta$ and then find angle $\beta$ by the sine law in triangle PQR
Edit: Fixed RP and RQ
A: 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! 
