# Finding side and angle of isosceles triangle inside two circles

I'm having a problem that I'm not sure how to solve (or if it's even possible). It's not homework, just something I'm struggling with for a project. :)

Basically, there are two circles, represented here by the red-ish lines, that have the same center point $E$. The radii of the circles are not known, these are the distances $AE$ and $CE$.

Distances $AE$ and $BE$ are identical.
Distances $CE$ and $DE$ are identical.
Hence, distances $AC$ and $BD$ are identical.

Known metrics:
$AC = BD = 10$ meters.
$AD = 200$ meters.

From $A$ and $B$ are tangent lines, marked as blue. From the tangent line that is connected by point $A$ to the line $AD$, the angle is $8^\circ$.

The problem here is: How do I find the radius of any of the circles? I'm thinking setting up an isosceles triangle is the way to move forward, but I'm stuck. Can the distance $CD$ and angle $CDE$ be found to solve the sides of the triangle? Is there another way of finding the radii that I haven't thought of?

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Let's calculate $CD$ using the Law of cosines in $\triangle CAD$: $$CD^2=10^2+200^2-2 \cdot 10 \cdot 200 \cdot \cos 82^{\circ} \Rightarrow$$ $$\Rightarrow CD \approx 198.855 \, \mathrm{m}.$$ Now using the Law of cosines in $\triangle ACD$, let's calculate $\cos \alpha$: $$200^2=10^2+CD^2-2 \cdot 10 \cdot CD \cdot \cos \alpha \Rightarrow$$ $$\Rightarrow \cos \alpha \approx -0.0896857.$$ Note that $\triangle CEF$ is a right-angled triangle, $CE=r$, and $CF=\frac{CD}{2}$, so: $$\cos(180^{\circ} - \alpha)= \frac{CF}{CE} \Rightarrow$$ $$\Rightarrow -\cos \alpha =\frac{CD}{2r} \Rightarrow$$ $$\Rightarrow r = \frac{CD}{-2\cos \alpha} \Rightarrow$$ $$\Rightarrow r \approx 1108.6 \, \mathrm{m}.$$