Animated Solution for Mathematics - Trigonometry: A polygon of nine sides, each of length 2, is inscribed in a circle. The radius of the circle is ..................

Welcome, future engineer. Today, we are not just solving a geometry problem; we are uncovering the hidden symmetry of a nonagon inscribed in a circle.

When you first look at a problem like this, it is easy to feel overwhelmed by the complexity of a nine-sided figure. But in the world of JEE Advanced, the secret to success is not brute force—it is the art of reduction. We must learn to see the simple within the complex.

Imagine a beautiful, symmetric circle. Inside this circle, we fit a regular polygon with exactly nine sides—a nonagon. Each side has a length of $s=2$.

If you try to analyze all nine sides at once, you will get lost. Instead, let us focus on a single 'slice' of this polygon. Imagine the center of the circle, point $O$, and two adjacent vertices of the polygon, $A$ and $B$.

By connecting $O$ to $A$ and $O$ to $B$, we create an isosceles triangle, $△OAB$. The sides $OA$ and $OB$ are both radii of the circle, which we call $R$. The base $AB$ is our polygon side of length $2$.

This triangle is our fundamental building block. Everything we need to know about the circle is encoded within this single, elegant shape.

Now, we need to relate the radius $R$ to the side length $2$. To do this, we need a right-angled triangle. We draw a perpendicular line from the center $O$ to the side $AB$, meeting it at point $L$.

Because $△OAB$ is isosceles, this line $OL$ is a miracle of geometry: it is the altitude, the median, and the angle bisector all at once. This means it splits the base $AB$ into two equal segments of length $1$, and it splits the central angle $\angle AOB$ into two equal angles of $\angle AOL=\frac{1}{2}\angle AOB$.

Since the total central angle of a full circle is $2\pi$ radians, and our nonagon divides this into $9$ equal parts, the total central angle is $\frac{2\pi }{9}$. Therefore, our half-angle $\angle AOL$ is exactly $\frac{\pi }{9}$.

We now have a perfect right-angled triangle, $△AOL$. We know the angle $\angle AOL=\frac{\pi }{9}$, the opposite side $AL=1$, and the hypotenuse $OA=R$.

The definition of the sine function is the ratio of the opposite side to the hypotenuse. Thus, we can write:

With a simple algebraic rearrangement, we find:

Recalling our trigonometric identities, we know that $\frac{1}{\sin (\theta )}=\csc (\theta )$. Therefore, the radius is simply $R=\csc \left(\frac{\pi }{9}\right)$.

This result is not just an answer; it is a general principle. For any regular polygon with $n$ sides of length $s$, the circumradius is:

By mastering this derivation, you have not just solved one problem—you have equipped yourself to handle any inscribed polygon problem that the JEE examiners can throw at you. Keep visualizing, keep simplifying, and keep pushing forward. You are doing great.