Of circles, tangents, and triangles?

Let $R$ be the radius of circle $A$ , and $r$ be the radius of circle $B$ . Since $\overline {CD} = 6r$ , it follows that $\overline {BD} = 3r$ . By Pythagorean Theorem, we can determine the length of the radius of circle $A$ , which would be essential in determining $\angle FAE$ .

$(BD)^{2} + (AB)^{2} = (AD)^{2}$

$(3r)^{2} + (R-r)^{2} = R^{2}$

$9r^{2} + R^{2} - 2Rr + r^{2} = R^{2}$

$10r^{2} = 2Rr$

$R = 5r$ .

By establishing that, we now draw a segment from $B$ to the point of tangency of $\overline {AF}$ . We let this point be $P$ . We can now get $\theta = \angle PAB$ . That is,

$sin (\theta) = \frac {BP}{AB} = \frac {r}{R-r}$

$sin (\theta) = \frac {1}{4}$

Our objective is to get $\angle FAE$ , which is equal to $2 \theta$ . Using the double angle identity for cosine, we get

$cos (2\theta) = 1 - 2sin^{2} (\theta)$

$cos (2 \theta) = 1 - 2( \frac {1}{4} )^{2}$

$cos (2 \theta) = 1 - \frac {1}{8}$

$cos (2 \theta) = \frac {7}{8}$

And so we find $\angle FAE = cos^{-1} (\frac{7}{8})$ .

using $a=7$ and $b=8$ , we find that $a \times b = \large \boxed {56 }$