We will work in argand plane. Let , $z = cos(\frac{\pi}{5}) + isin(\frac{\pi}{5})$. Let , $A_{i} = z^{i-1}$ , $\forall i \in (1,2,...,10)$ Now see that , $A =\frac{z^3(z+z^4)-z^5(1+z^3)}{z^3-z^5} = \frac{z(z^3+1)}{z+1}$ $B = 0$ $C =\frac{z^8(z+z^9)-z^{10}(1+z^8)}{z^8-z^{10}}=\frac{z(z^8+1)}{z+1}$ Now, Let $A' = 0,B'=\frac{1-z^2}{2} , C' = 1$ it's easy to see that $\triangle ABC \sim \triangle A'B'C'$ See that $1-z^2 = {1 - cos(\frac{2\pi}{5}) - isin(\frac{2\pi}{5})} = -2isin(\frac{\pi}{5})z$ $1+z^2= {1 + cos(\frac{2\pi}{5})+isin(\frac{2\pi}{5})} = 2cos(\frac{\pi}{5})z$ Now, $\angle B'A'C' = arg (-z) + arg (isin(\frac{\pi}{5})) = \frac{\pi}{2} - \frac{\pi}{5} = \frac{3\pi}{10}$ $\angle A'B'C' = arg (\frac{z^2-1}{z^2+1}) = arg(itan\frac{\pi}{5}) = \frac{\pi}{2}$ $\angle A'C'B' = \frac{\pi}{5}$ So angles of $\triangle ABC$ are $\frac{3\pi}{10},\frac{\pi}{2},\frac{\pi}{5}$