Seven circles inscribed in a circle problem This is the detail given, Diameter of inner circle is 50. I need to find out middle circle and outer circle radius 
The answer is given at the end of the book but I want to learn how to do the calculation. I know that outer circle is 126,6 and smaller middle cirles are 38.3 .
I have tried solving it this way: Having the equilateral triangle, the height of it is 43.3. Doubling that is 86,6. Equilateral height line doubled in dark red.
Having the equilateral triangle, the height of it is 43.3. Doubling that is 86,6. Equilateral height line doubled in dark red. I have been stuck at this for hours, so I need tips what to do next.
 A: Let $O$ be inner circle center, and $A$ and $B$ be two adjacent middle circles centers. Then we know $\angle AOB = 2\pi/7$ as there are $7$ middle circles around and all their adjacent center's angles on $O$ will be same. We know inner circle's radius is $50$, let r be the middle circle's radius. Then the triangle given in here  will have sides $50+r, 50+r$ and $2r$ (not an equilateral triangle as in your question) We can draw the angle bisector of $\angle AOB$, and applying $\sin(\pi/7) = \frac{r}{50+r}$ in the right angled triangle which will form, we get $r=\frac{50\sin(\pi/7)}{1-\sin(\pi/7)}\approx 38.32108$. Now radius of outer circle is just $50 + 2r \approx 126.64216 $
A: It is enough to consider the triangle $\triangle {OAB}$ involving the inner circle of radius $50$ and two of the searched seven circles of unknown radius $r$. Since  angle $\angle{BOC}=\dfrac{\pi}{7}$ we have the equation $$\frac{r}{50+r}=\sin\left(\dfrac{\pi}{7}\right)=0.433883739$$ so $r=38.32108076$. From which the radius of the outer circle is equal to $126.64216153$

