7 doesn't seem nice?

The angles of triangle are $A=\frac{\pi}{7},B=\frac{2\pi}{7}, C=\frac{4\pi}{7}$ . Using the usual convention, $a,b,c$ are the sides opposite to $A,B,C$ respectively, $\Delta$ is area of $\Delta ABC$ and $R$ is the circumradius of $\Delta ABC$ .

$\Delta=\frac{abc}{4R}$

Using Sine Rule , we get $\Delta=2R^2\sin A\sin B\sin C=2\sin A\sin B\sin C$

Consider $x=\frac{\pi}{7}$ .

$7x=\pi\Rightarrow \tan 7x=0$

$\frac{7\tan x-35\tan^3x+21\tan^5x-tan^7x}{1-21\tan^2x+35\tan^4x-7\tan^6x}=0$

$\Rightarrow 7\tan x-35\tan^3x+21\tan^5x-tan^7x=0$

Since we know, $\tan x\ne 0$ ,

$7-35\tan^2x+21\tan^4x-tan^6x=0$

Clearly, we have an equation of degree $6$ with roots as $\tan \frac{n\pi}{7}, n\in \{1,2,3,4,5,6\}$

Replacing $\tan^2x$ by $p$ ,

\(p^3-21p^2+35p-7=0)

Now, here's an equation of degree \(3\) with roots as $\tan^2\frac{n\pi}{7}, n\in \{1,2,3\}$

Thus, $\tan^2\frac{\pi}{7}\tan^2\frac{2\pi}{7}\tan^2\frac{4\pi}{7}=7$

Getting back to our triangle,

$\Delta=2\sin \frac{\pi}{7}\sin \frac{2\pi}{7}\sin \frac{4\pi}{7}\\=2\tan \frac{\pi}{7}\tan \frac{2\pi}{7}\tan \frac{4\pi}{7}\cos \frac{\pi}{7}\cos \frac{2\pi}{7}\cos \frac{4\pi}{7}\\=2\times (-\sqrt{7}) \times \cos \frac{\pi}{7}\cos \frac{2\pi}{7}\cos \frac{4\pi}{7}$

Multiplying and Dividing by $\sin\frac{\pi}{7}$ ,

$\Delta=2\times (-\sqrt{7}) \times \sin\frac{\pi}{7}\cos \frac{\pi}{7}\cos \frac{2\pi}{7}\cos \frac{4\pi}{7}\times \frac{1}{\sin\frac{\pi}{7}}\\=-2\sqrt{7}\times\frac{1}{8}\times\frac{\sin\frac{8\pi}{7}}{\sin\frac{\pi}{7}}=\frac{\sqrt{7}}{4}$

So finally, $a=7,b=4$ , and $7^2+4^2=\boxed{65}$

To make the typing a bit easier, we let $t=\pi/7$ throughout.

By the Law of Sines, the sides are $2\sin{t}, 2\sin{2t}, 2\sin{4t}$ , and the area is $2(\sin{t})(\sin{2t})(\sin{4t})=\frac{\sqrt{7}}{4}$ , so that $a^2+b^2=\boxed{65}$ . We have used the identity (for odd $n$ ) $\prod_{k=1}^{(n-1)/2}\sin\left(\frac{k\pi}{n}\right)=\frac{\sqrt{n}}{2^{(n-1)/2}}$ for $n=7$ , keeping in mind that $\sin\left(\frac{3\pi}{7}\right)=\sin\left(\frac{4\pi}{7}\right)$ .

∆=abc/4R a=2RsinA,b=2RsinB,c=2RsinC So ∆=2R^2 sinAsinBsinC angles are π÷7,2π÷7,4π÷7. R=1....

Apply $\sqrt{s (s - a)(s - b)(s- c)}$ for area wanted.

Only $2 R$ = $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$ works but not $\Delta = \frac12 R^2$ ( $sin 2 A + \sin 2 B + \sin 2 C).$

$x + 2 x + 4 x = \pi$

$\implies x = \frac{\pi}{7}$

Since $R = 1,$

$a = 2 \sin \frac{\pi}{7}$ = 0.86776747823511+

$b = 2 \sin \frac{2 \pi}{7}$ = 1.5636629649360+

$c = 2 \sin \frac{4 \pi}{7}$ = 1.9498558243636+

$s$ = $0.5 (a + b + c)$ = 2.1906431337674+

Area wanted, $\Delta$ = 0.66143782776614+

$\Delta^2$ = $\frac{7}{16} \implies \Delta = \frac{\sqrt7}{4}$

Redefine $a$ and $b$ here:

$a^2 + b^2$ = 49 + 16 = 65

Answer: $\boxed{65}$

More info including already mentioned identity: $sin(\pi/7)sin((2\pi)/7)sin((3\pi)/7) = \frac{\sqrt{7}}{8}$

can be found here here

It is easy to show that : $\displaystyle{\Delta =\cfrac { abc }{ 4R } \\ \Rightarrow \Delta =2\sin { \cfrac { \pi }{ 7 } } \sin { \cfrac { 2\pi }{ 7 } } \sin { \cfrac { 3\pi }{ 7 } } }$

Now I recently studied a note based on complex number Techniques , By Deepanshu gupta , there i found exactly same series , which he solved by Complex numbers . So I got answer without much effort. That's why I really loved this website , I found very interesting things to learn... :)

Here It is Complex Numbers Techniques

7x = 180, so the smallest angle x = 180/7.

Area ΔABC = ½ ab sin ϒ. a/sinα = b/sinβ = 2R, so
Area ΔABC = 2R2 sin α .sinβ .sin ϒ. = 2(sin (180/7) .sin(360/7) .sin (720/7) =0.6614378