Let $a$ , $b$ , $c$ be the three sides of the triangle and $R$ its circumradius. Denote the requested product by $P$ .

In this solution we will use the following trigonometric identities:

• Double angle formula $\sin \left( 2x \right)=2\sin x \cos x$

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• Extended Sine Rule $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=2R$

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• $\sin A+\sin B+\sin C=4\cos \dfrac{A}{2}\cos \dfrac{B}{2}\cos \dfrac{C}{2} \ \ \ \ \ \ \ (1)$

Here we go: $\begin{aligned} P & =100\dfrac{\left( \sin \dfrac{A}{2}\cos \dfrac{A}{2} \right)\left( \sin \dfrac{B}{2}\cos \dfrac{B}{2} \right)\left( \sin \dfrac{C}{2}\cos \dfrac{C}{2} \right)}{{{\left( \cos \dfrac{A}{2}\cos \dfrac{B}{2}\cos \dfrac{C}{2} \right)}^{2}}} \\ & =100\dfrac{\left( \dfrac{1}{2}\sin A \right)\left( \dfrac{1}{2}\sin B \right)\left( \dfrac{1}{2}\sin C \right)}{\dfrac{1}{16}{{\left( \sin A+\sin B+\sin C \right)}^{2}}} \\ & =100\dfrac{\dfrac{1}{8}\left( \dfrac{a}{2R} \right)\left( \dfrac{b}{2R} \right)\left( \dfrac{c}{2R} \right)}{\dfrac{1}{16}{{\left( \dfrac{a}{2R}+\dfrac{b}{2R}+\dfrac{c}{2R} \right)}^{2}}} \\ & =100\dfrac{2\dfrac{1}{2{{R}^{2}}}\dfrac{abc}{4R}}{{{\left( \dfrac{a+b+c}{2R} \right)}^{2}}} \\ & =100\dfrac{\dfrac{1}{\cancel{{{R}^{2}}}}\left[ Area\text{ of }ABC \right]}{\dfrac{{{\left( Perimeter\text{ of }ABC \right)}^{2}}}{4\ \cancel{{{R}^{2}}}}} \\ & =400\dfrac{18}{{{20}^{2}}} \\ & =\boxed{18} \\ \end{aligned}$

Note : In fact the value of the circumradius is not needed.

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Proof of $(1)$ $\begin{aligned} \sin A+\sin B+\sin C & =2\sin \dfrac{A+B}{2}\cos \dfrac{A-B}{2}+2\sin \dfrac{C}{2}\cos \dfrac{C}{2} \\ & =2\cos \dfrac{C}{2}\cos \dfrac{A-B}{2}+2\cos \dfrac{A+B}{2}\cos \dfrac{C}{2} \\ & =2\cos \dfrac{C}{2}\left( \cos \dfrac{A-B}{2}+\cos \dfrac{A+B}{2} \right) \\ & =2\cos \dfrac{C}{2}\left( 2\cos \dfrac{\dfrac{A-B}{2}+\dfrac{A+B}{2}}{2}\cos \dfrac{\dfrac{A-B}{2}-\dfrac{A+B}{2}}{2} \right) \\ & =4\cos \dfrac{A}{2}\cos \dfrac{B}{2}\cos \dfrac{C}{2} \\ \end{aligned}$

From here , we gather that $\large \tan \left( \dfrac A2 \right) = \sqrt{\dfrac{(s-b)(s-c)}{s(s-a)} }, \quad \tan \left( \dfrac B2 \right) = \sqrt{\dfrac{(s-a)(s-c)}{s(s-b)} }, \quad \tan \left( \dfrac C2 \right) = \sqrt{\dfrac{(s-a)(s-b)}{s(s-c)} }$

where $s$ is the semiperimeter.

Let $P$ denote the product in question, then $\begin{array} {r c l} \left( \dfrac P{100} \right)^2 &=& \tan^2 \left( \dfrac A2 \right) \tan^2 \left( \dfrac B2 \right) \tan^2 \left( \dfrac C2 \right)\\ \phantom0 \\ &=& \dfrac{(s-b)(s-c)}{s(s-a)} \cdot \dfrac{(s-a)(s-c)}{s(s-b)} \cdot \dfrac{(s-a)(s-b)}{s(s-c)} \\ \phantom0 \\ &=& \dfrac{ (s-a)(s-b)(s-c)}{s^3} \\ \phantom0 \\ &=& \dfrac{ s(s-a)(s-b)(s-c)}{s^4} \\ \phantom0 \\ &=& \dfrac{ 18^2}{10^4} \\ \phantom0 \\ P &= & \boxed{18} \end{array}$

The penultimate step follows from Heron's formula .

I did it a bit differently:P

From the previous problem's solutions I got to know that the side lengths are : $\boxed{\dfrac{39.56}{5}}$ , $\boxed{\dfrac{28.44}{5}}$ , $\boxed{\dfrac{32}{5}}$

Looking closely at them I realised that the angles should not vary much from one another

Assuming that their values vary by 5°,

And since A+B+C=180°(by angle sum property of a triangle)

I found the values $\boxed{55,60,65}$ to work

Then I used the calculator to calculate the values of $\dfrac{tan55°}{2}$ , $\dfrac{tan65°}{2}$

Bcz I knew that tan30°= $\dfrac{1}{√3}$

And after multiplying these values I got, 18.6-18.9≈ $\boxed{18}$ ,which is the correct answer:)

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Alternate approach as suggested by @Pi Han Goh :

diagram

Using the law of cotangents,

${\dfrac{\cot \left({\tfrac{\alpha}{2}}\right)}{s-a}}$ = ${\dfrac{\cot \left({\tfrac{\beta}{2}}\right)}{s-b}}$ = ${\dfrac{\cot \left({\tfrac{\gamma}{2}}\right)}{s-c}}$ = $\dfrac{1}{r}$

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$\textcolor{#3D99F6}{\tan{\tfrac{A}{2}}}$ = $\textcolor{forestgreen}{\dfrac{r}{s-a}}$

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$\textcolor{#3D99F6}{\tan{\tfrac{B}{2}}}$ = $\textcolor{forestgreen}{\dfrac{r}{s-b}}$

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$\textcolor{#3D99F6}{\tan{\tfrac{C}{2}}}$ = $\textcolor{forestgreen}{\dfrac{r}{s-c}}$

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Where r is the circumradius= 4 and s is the semi-perimeter=10 Plugging in the values of s ,r ,a ,b ,and c we get,

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$\textcolor{#3D99F6}{\tan{\tfrac{A}{2}}}$ = $\textcolor{forestgreen}{\dfrac{20}{50-39.56}}$

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$\textcolor{#3D99F6}{\tan{\tfrac{B}{2}}}$ = $\textcolor{forestgreen}{\dfrac{20}{50-28.44}}$

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$\textcolor{#3D99F6}{\tan{\tfrac{C}{2}}}$ = $\textcolor{forestgreen}{\dfrac{20}{50-32}}$

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$=>\tan{\dfrac{A}{2}} = 1.91712278$

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$=>\tan{\dfrac{B}{2}} =0.927352159$

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$=>\tan{\dfrac{C}{2}} = 1.111111$

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incomplete