If  ${\sin }^2\left(10°\right)\sin \left(20°\right)\sin \left(40°\right)\sin \left(50°\right)\sin \left(70°\right)=\alpha -\frac{1}{16}\sin \left(10°\right)$, then $16+{\alpha }^{-1}$ is equal to

$\begin{array}{l}\sin {10}^{\circ }\left(\frac{1}{2}\cdot 2\sin {20}^{\circ }\sin {40}^{\circ }\right)\cdot \sin {10}^{\circ }\sin \left({60}^{\circ }-{10}^{\circ }\right)\sin \left({60}^{\circ }+{10}^{\circ }\right)\\ \sin {10}^{\circ }\frac{1}{2}\left(\cos {20}^{\circ }-\cos {60}^{\circ }\right)\cdot \frac{1}{4}\sin {30}^{\circ }\\ \frac{1}{2}\cdot \frac{1}{4}\cdot \frac{1}{2}\cdot \sin {10}^{\circ }\left(\cos {20}^{\circ }-\frac{1}{2}\right)\\ =\frac{1}{32}\left(2\sin {10}^{\circ }\cos {20}^{\circ }-\sin {10}^{\circ }\right)\\ =\frac{1}{32}\left(\sin {30}^{\circ }-\sin {10}^{\circ }-\sin {10}^{\circ }\right)\\ =\frac{1}{32}\left(\frac{1}{2}-2\sin {10}^{\circ }\right)\\ =\frac{1}{64}\left(1-4\sin {10}^{\circ }\right)\\ =\frac{1}{64}-\frac{1}{16}\sin {10}^{\circ }\\ \text{ Hence }\alpha =\frac{1}{64}\Rightarrow 16+{\alpha }^{-1}=80\end{array}$