3 Circles & 1 Triangle

• Radius B -> R=sqrt(2)*r //r=radius of A
• Radius C-> |R=sqrt(2)*2r
• cos A= (AB^2 + AC^2 - BC^2) / 2(AB)(AC)
• where AB=r+R, BC=R+|R, AC=r+|R :)

If the ratio of the circular areas for $A,B,C$ are $1:2:8$ , then we have radii lengths $r_{A} = R, r_{B} = \sqrt{2} R, r_{C} = 2\sqrt{2}R$ . We can calculate the cosine of $\angle A$ by the Law of Cosines:

$\cos \angle A = \frac{AB^{2} + AC^{2} - BC^{2}}{2(AB)(AC)} = \frac{(1+\sqrt{2})^{2}R^{2} + (1 + 2\sqrt{2})^{2}R^{2} - (3\sqrt{2})^{2}R^{2}}{2(1+\sqrt{2})(1+2\sqrt{2})R^{2}} = \frac{3\sqrt{2} - 3}{3\sqrt{2}+5} \cdot \frac{3\sqrt{2}-5}{3\sqrt{2}-5} = \frac{15+ 18 - 24\sqrt{2}}{-7} = \boxed{\frac{24\sqrt{2}-33}{7}}.$