# given an altitude and a viewing angle how do I determine the distance of the viewing area

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).

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Here is the situation as it seems to be described.

$\hspace{1cm}$

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}

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in this drawing I need to know that arc length that is between where h intersects the earth and where c intersects. so we now know c and h but we need the length of the last side. – Mike LP Oct 8 '12 at 20:56
It would be helpful if you would explain what you want. It is hard to tell which angle is $30^\circ$ and which distance you want. – robjohn Oct 8 '12 at 21:53
in this example your theta is 30 deg – Mike LP Oct 8 '12 at 21:56
If I had your skills at illustrating the problem it would have been very clear. It's an amazing rendering of the problem – Mike LP Oct 8 '12 at 21:57
this is exactly what I was looking for. Thank you so much! – Mike LP Oct 8 '12 at 23:38

Let $R$ be the radius of the Earth, and $h$ the height of the tower. Let $A$ be the point of attachment of the cable, $B$ the top of the tower, and $C$ the centre of the Earth.

If $\theta=\angle CAB$, by the Sine Law $$\frac{\sin\theta}{R+h}=\frac{\sin 30^\circ}{R}=\frac{1}{2R}.$$ It follows that $$\sin\theta=\frac{1}{2}+\frac{h}{2R}.$$ With your value of $R$, and $h=100$, the calculator gives $\theta\approx 149.4794$ degrees.

Now we can calculate the angle at $C$ (about $0.520596$ degrees), and hence the distance, along the surface of the Earth, from the bottom of the tower to the point of attachment.

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Wouldn't $\theta$ actually be $180^\circ-30.520596^\circ=149.479404^\circ$? – robjohn Oct 9 '12 at 0:03
Yes, thanks, will correct. – André Nicolas Oct 9 '12 at 0:11

Draw a triangle from the center of the earth (A) to the top of the tower (B) to a point at the edge of visibility (C). We are given that $\angle B=30^\circ$. Let $R$ be the radius of the earth. The slant range, $BC$ can be found from the cosine law: $AC^2=R^2=AB^2+BC^2-2AB\cdot BC\cos 30^\circ =(R+100)^2+BC^2-2BC(R+100)\frac {\sqrt 3}2$

where it would be good to cancel the $R^2$ terms before using your calculator to avoid loss of precision. Having found $BC$ you can find the angle at the center of the earth, then the arc on the surface of the earth if that is the way you want to express it. Even at 100 km, I suspect the earth radius will not change the answer much.

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thanks, the actual subject is approx 400km high. – Mike LP Oct 8 '12 at 15:25