Circles and a... Triangle

The ratio of the big circle's circumference to the small circle's circumference is really just the ratio of the big circle's diameter to the small circle's diameter (because $\dfrac{\pi\cdot D_{big}}{\pi\cdot D_{small}}$ reduces to $\dfrac{D_{big}}{D_{small}}$ ). To make it even easier, we can just find the ratio of the big circle's radius to the smaller circle's radius, or $\dfrac{r_{big}}{r_{small}}$ (we can simply multiply above and below by $\frac{1}{2}$ ).

If we draw a line from the centers of the circles to the top left vertex of the triangle, we have a right triangle as in the diagram to the right. Because 2 of this triangle's sides are the radii of the circles, we can use a trigonometric function to describe their ratio.

If we use the angle labeled $\theta$ in the diagram, the ratio of the big circle's radius to the small circle's radius turns out to be $\csc(\theta)$ , as the problem asks for. Now all we have to find is the value of $\theta$ . Because this triangle is equilateral, each angle is $60^\circ$ , and because $\theta$ is half of this, our final answer will be $\boxed{30^\circ}$

Simple, just join the center of the small circle to the vertex of the equilateral triangle.

It becomes a right angled triangle.

The angle formed at the vertex is $30^\circ$

Using basic Trigonometry, we have $\sin30^\circ = \dfrac{3}{x}$ , where $x$ is the radius of the bigger circle as well as the hypotenuse.

$\therefore$ $x=6$

Now, we know both the circles' radii, i.e. $3$ and $6$ .

Required ratio = $cosec\theta = \dfrac{2\pi(6)}{2\pi(3)} = \dfrac{6}{3} = 2$

$\implies$ $\theta = \boxed{30^\circ}$