Hello, shane99!

A circle with center $\displaystyle O$ intersects another circle with center $\displaystyle P$ at points $\displaystyle C$ and $\displaystyle D.$

If $\displaystyle \angle COD = 30^o$ and $\displaystyle \angle CPD = 60^o,$

what is the ratio of the area of the circle $\displaystyle O$ to the area of the circle $\displaystyle P$?

Answer: .$\displaystyle 1: (2-\sqrt{3})$ Here's the diagram ... without the circles.

Code:

C
*
R * | *
* | * r
* | *
O * 30° d| 60° * P
* | *
* | * r
R * | *
*
D

The radius of circle $\displaystyle O$ is $\displaystyle R.$

The radius of circle $\displaystyle P$ is $\displaystyle r.$

The ratio of the areas is: .$\displaystyle \text{Ratio} \:=\:\frac{\pi R^2}{\pi r^2} \;=\;\frac{R^2}{r^2}$

Apply the Law of Cosines to $\displaystyle \Delta OCD$ and $\displaystyle \Delta PCD:$

. . $\displaystyle R^2 + R^2 - 2R^2\cos30^o \:=\:d^2 \:=\:r^2 + r^2 - 2r^2\cos60^o$

. . $\displaystyle 2R^2(1-\cos30^o) \;=\;2r^2(1-\cos60^o) \quad\Rightarrow\quad \frac{R^2}{r^2} \;=\;\frac{1-\cos60^o}{1-\cos30^o} $

Therefore: .$\displaystyle \text{Ratio} \;=\;\frac{R^2}{r^2} \;=\;\frac{1 - \frac{1}{2}}{1-\frac{\sqrt{3}}{2}} \;=\;\frac{2}{2}\cdot\frac{1-\frac{1}{2}}{1-\frac{\sqrt{3}}{2}} \;=\; \frac{1}{2-\sqrt{3}} $