10.143. A common chord of two circles subtends arcs of $60^{\circ}$ and $120^{\circ}$. Find the ratio of the areas of these circles.
## Solution.

Let $AB$ be the common chord, $R_{1}, R_{2}$ be the radii of the respective circles (Fig. 10.128). By the cosine theorem,

$A B^{2}=R_{1}^{2}+R_{1}^{2}-2 R_{1} R_{1} \cos 60^{\circ}=R_{1}^{2}$.

On the other hand, $A B^{2}=R_{2}^{2}+R_{2}^{2}-2 R_{2} R_{2} \cos 120^{\circ}=3 R_{2}^{2}$. Therefore, $R_{1}^{2}=3 R_{2}^{2}$. The areas of the circles are $S_{1}=\pi R_{1}^{2}, S_{2}=\pi R_{2}^{2}$.

The ratio $\frac{S_{1}}{S_{2}}=\frac{3 R_{2}^{2} \pi}{\pi R_{2}^{2}}=\frac{3}{1}$.

Answer: $3: 1$.