The cone and the triangle

$AB = AC$ , since in a hyperbola, if $c$ is distance to the focus, and $a$ and $b$ are the half lengths of the axes of the hyperbola, $a^2 + b^2 = c^2$ . So then I have drawn an isosceles triangle in the photograph since $c = 1 = c^2$ . Then, the problem becomes relatively straightforward. We know that the slope of the right asymptote is 2016, and $\tan \theta = \dfrac{y}{x} \text{ or in this case, } \dfrac{b}{a} = 2016$ , so we know that $\theta = \arctan{2016}$ . Because we have an isosceles triangle, and we are asking for the an angle in the base of this triangle, use wolfram alpha to compute $\left \lfloor 10000 * \delta \right \rfloor$ , where $\delta = \dfrac{180^{\circ} - \arctan{2016}^{\circ} }{2}$ , giving an answer of $\boxed{892148}$ .