In $\Delta ABC$, if $$\tan{\dfrac{A}{2}}\tan{\dfrac{B}{2}}=\dfrac{1}{2}\\\sin{\dfrac{A}{2}}\sin{\dfrac{B}{2}}\sin{\dfrac{C}{2}}=\dfrac{1}{10}$$
Find the $\angle C$
My try: since $$2\sin{\dfrac{A}{2}}\sin{\dfrac{B}{2}}=\cos{\dfrac{A}{2}}\cos{\dfrac{B}{2}}$$ $$\left(\cos{(\dfrac{A-B}{2})}-\cos{(\dfrac{A+B}{2})}\right)\sin{\dfrac{C}{2}}=\dfrac{1}{5}$$ since $$\cos{(\dfrac{A+B}{2})}=\sin{\dfrac{C}{2}}$$
so $$\left(\cos{(\dfrac{A-B}{2})}-\sin{\dfrac{C}{2}}\right)\sin{\dfrac{C}{2}}=\dfrac{1}{5}$$ Then I can't go on.
Denote $~~a=\tan\dfrac{A}{2}$, $~~b=\tan\dfrac{B}{2}$ $($let $a\le b$$)$.
$\sin\dfrac{A}{2}\sin\dfrac{B}{2}\sin\dfrac{C}{2}=\dfrac{1}{10}$;
$\sqrt{\dfrac{1-\cos A}{2}} \cdot \sqrt{\dfrac{1-\cos B}{2}} \cdot \sqrt{\dfrac{1-\cos C}{2}} = \dfrac{1}{10}$;
$(1-\cos A) (1-\cos B)(1-\cos C) = \dfrac{2}{25}$;
$(1-\cos A) ~ (1-\cos B) ~ (1+\cos(A+B)) = \dfrac{2}{25}$;
$\dfrac{2a^2}{1+a^2} \cdot \dfrac{2b^2}{1+b^2} \cdot \dfrac{2(1-ab)^2}{(1+a^2)(1+b^2)} = \dfrac{2}{25}$;
$\dfrac{ab|1-ab|}{(1+a^2)(1+b^2)}=\dfrac{1}{10}$.
Now we get system:
$$ \left\{ \begin{array}{l} ab=\dfrac{1}{2};\\ \dfrac{ab|1-ab|}{(1+a^2)(1+b^2)}=\dfrac{1}{10}. \end{array} \right. $$
$b=\dfrac{1}{2a}$ $\implies$ $\dfrac{a^2}{(1+a^2)(4a^2+1)}=\dfrac{1}{10}$ ;
$4a^4-5a^2+1=0$;
$(2a-1)(2a+1)(a-1)(a+1)=0$.
there are $2$ positive roots for this eq.:
$a=\dfrac{1}{2}$ and $a=1$.
If $a=\dfrac{1}{2}$, then $b=1$, then $A = \arctan\dfrac{4}{3}=\arcsin\dfrac{4}{5}=\arccos\dfrac{3}{5}$, $~B=\pi/2$, $~C=...$;
If $a=1$, then $b=\dfrac{1}{2}$ (simply permutation of $a,b$).
(don't know if this is the simplest way, but at least this is $\approx$ clear :)
Finally,
$C=\arctan\dfrac{3}{4}=\arcsin\dfrac{3}{5}=\arccos\dfrac{4}{5} = 0.643501108793...$.