## Tuesday, June 16, 2015

### Cloudy Puzzle: Solution

We want to establish a mathematical relationship between the key variables of the problem: $h, R, \theta$ and $\phi$.

Consider the triangle PRQ; we can write, $\tan\left(\frac{\pi}{2} - \theta \right) = \frac{QP}{RQ} = \frac{QP}{OQ - OR}.$
Recognizing that the length of the segment OP is $R+ h$, we can further write:
$\tan\left(\frac{\pi}{2} - \theta \right) = \frac{(R+h) \cos \phi}{(R+h) \sin \phi - R}.$
Using the variable $a = h/R$, we can try to solve for $\phi$, which in this case leads to a quadratic equation. Let us limit $\theta < \pi/2$, since all other cases are trivial extensions.

If we let,  $A = a^{2} \tan^{2} \theta + a^{2} + 2 a \tan^{2}\theta + 2 a + \tan^{2} \theta,$ and $B = a \tan \theta + \tan \theta - 1,$ then the two solutions are: $\phi = -2 \tan^{-1} \left( \frac{a + 1 \pm \sqrt{A}}{B} \right).$
Due to the geometry, acceptable values of $\theta$ and $\phi$ lie in the region $[0, \pi/2]$. This constraint may be used to cull out the extraneous root.

It is useful to consider the limit of $a = h/R \ll 1.$ If we take the limit as it goes to zero, then  $\phi = -2 \tan^{-1} \left( \frac{ \pm|\tan \theta| + 1}{\tan \theta - 1} \right).$