What is the formula to solve this problem.
Say I have a tower 100km tall and I want to determine the distance from the base of the tower to where a cable is attached to the ground. The cable forms a $30^\circ$ angle with the tower and starts at the top of the tower. We do not know the length of the cable.
This isn't a simple right angle problem because the altitude is great enough that the earth's radius must be factored in.
All we really know is the top of the tower is 100 km high, the angle between the cable and the tower is $30^\circ$, and the earth's radius is 6371km (lets assume a perfect circle).
Best Answer
Here is the situation as it seems to be described.
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The Law of Sines says that $$ \sin(\phi)=\frac{r+h}{r}\sin(\theta)\tag{1} $$ The two values of $\phi$ specified in $(1)$ correspond to the two points of intersection of the line with the circle. The one pictured above is $$ \phi=\pi-\arcsin\left(\frac{r+h}{r}\sin(\theta)\right)\tag{2} $$ The arclength from the base of the tower to the ground point of the cable is $$ \begin{align} r(\pi-\theta-\phi) &=r\left(\arcsin\left(\frac{r+h}{r}\sin(\theta)\right)-\theta\right)\\ &=6371\left(\arcsin\left(\frac{6371+100}{6371}\frac12\right)-\frac\pi6\right)\\ &=57.89\text{ km}\tag{3} \end{align} $$ where the angles in $(3)$ are given in radians.
Just as a point of comparison, the answer without considering the curvature of the earth would be $$ \begin{align} h\tan(\theta) &=100\frac1{\sqrt{3}}\\ &=57.74\text{ km}\tag{4} \end{align} $$
Take it to the Limit
The formula in $(3)$ becomes the formula in $(4)$ when $r\to\infty$: $$ \begin{align} \lim_{r\to\infty}r\left(\arcsin\left(\frac{r+h}{r}\sin(\theta)\right)-\theta\right) &=h\lim_{r\to\infty}\frac{\arcsin\left(\left(1+\frac hr\right)\sin(\theta)\right)-\theta}{\frac hr}\\ &=h\lim_{t\to0}\frac{\arcsin((1+t)\sin(\theta))-\theta}{t}\\ &=h\lim_{t\to0}\frac{\sin(\theta)}{\sqrt{1-(1+t)^2\sin^2(\theta)}}\\ &=h\tan(\theta) \end{align} $$