So long radio song

Draw a line from the top of the tower to the center of the Earth. This is length R + 2 km, where R is the radius of the Earth.

Draw a second line from the right most point of X to the center of the Earth. This is length R and it makes a right angle with the line connecting the tower and the rightmost point of X.

The angle between these two lines is $\theta$ . Therefore, $\cos \theta = \frac {R}{R + 2 km}$ . Using R = 6370 km, we have $\cos \theta = \frac {6370 km}{6372 km}$ . $\theta$ = .0251 radians.

X is the arc length that \theta subtends, therefore X = R $\theta$ . X = 6370 km * .0251 radians X = 160 km

The length of an arc $X$ of a circular sector with radius $R$ and angle $\theta$ is $$X=R\theta$$ Consider the right triangle shown in the figure below. The cosine of the angle $\theta$ is given by basic trigonometry to be $$\cos \theta = \frac{R}{R+h}$$ $$\Rightarrow \theta = \cos^{-1} \left( {\frac{R}{R+h}} \right)$$ So the arc length is $$X=R \, \cos^{-1} \left( {\frac{R}{R+h}} \right)$$ $$=\left( 6370 \ \mathrm{km} \right) \, \cos^{-1} \left( {\frac{6370 \ \mathrm{km}}{6372 \ \mathrm{km}}} \right)$$ $$=\fbox{159.6 km}$$ Triángulo rectángulo

What we need to know is the arclength lying between the base of the FM antenna and the point where the line of sight is lost. The line of sight is by definition the tangent that passes through the antenna top. If O is the center of the earth, B is the base of the antenna, A the antenna top, C the common point of the tangent line and the circle and φ=<AOC we have that the triangle AOC is right (<ACO=π/2) and therefore:

$cosφ=\frac{R}{R+h}$

$x=φ\cdot{R}$

where x is the arc length A, R is the radius of the earth and h the height of the antenna.The solution is

$x=arccos(\frac{R}{R+h})\cdot{R}=159.6{ km}$

To find how far away you can be from the tower before you lose line of sight to it , knowing that the Earth is curved , we draw a tangent line from the tower to the circle (that is earth) , naming that point A . Now the radius of the earth forms an angle that has 90 degrees with that point . To find our x ,we first need to find the angle ∠TOA(T being the highest point of the tower ,and O being the center of the earth .
cos(∠TOA) = 6370/(6370+2) => ∠TOA= arccos(6370/6372) => ∠TOA= 1.435575 degrees . To find the x we use the next formula : x= (∠TOA * π /180) * 6370 => x= 159.6 Km .

This problem involves basic geometry/trigonometry. Take a look at the image below:

Alt text

Here $AB$ is the radio tower. So $AB=2 \text{km}$ .

$O$ is the center of the earth. So $OB=OC=6370\text{km})$ .

And $\stackrel{\frown}{BC}=X$ .

Notice that $\angle AOC=\cos^{-1}(\frac{OC}{OA})=\cos^{-1}(\frac{6370}{6372})=0.025055\text{radians}$ .

And $X=OB\times \angle AOC (\text{in radians})=159.6 \text{km} [ \text{approximately}]$ .

This problem isn't too difficult if you can find the "right" triangle (haha). Let the center of the circle be point O, the top of the tower point A, and the farthest point with a line of sight to the tower point B. $\overline{AB}$ is perpendicular to $\overline{OA}$ because a tangent line to a point on a circle is perpendicular to the radius to that point. Now, we have the right triangle $\triangle{OAB}$ . OA=6370+2=6372 and OB=6370, so to find AB, we use the Pythagorean theorem and get AB=159.64 km.

assuming the earth to be a sphere or radius R=6370 km.

Let angle subtended at the center of earth by arc X = A degrees. Then Cos(A)=R/(R+2) =6370/6372=0.99968. A=1.435575 degrees(approx) therefore distance X=(2.pi.R.A)/360 = 159.60 (approx).

the angle between line joining center of earth and tower and perpendicular drawn from center to the horizontal line can be found by arccos function. this multiplied by radius gives the value of 'x'.

The line from the mast to the Earth will be a tangent to the Earth. This means the radius will be at a right angle to where the tangent touches the Earth. We can therefore form a right triangle with sides of length radius and radius + 2 or 6370km and 6372km with 6372km being the hypotenuse. To find the length of x we need to find the angle made at the centre of the Earth from this triangle. So if this angle is theta (θ) (in radians). $\cos \theta = \frac{6370km}{6372km}$ $\theta = \cos^{-1}\frac{6370}{6372}$ $\theta = 0.025056$ X is the arc of a circle so its length will be given by the radius multiplied by the angle the arc makes of the circle (in radians). $x=r\theta$ $x=159.6km$

Basically we have to find the point on a circle from where if you draw a tangent, then it cuts the point at the tip of the tower. So let's get started by setting up our co-ordinate system. Let the point on the top of the tower be on the $y$ axis. Let the circle/sphere (earth), be centered at origin and call the radius $r$ . We'll substitute the values later. Now, the point on the top of the tower must be $(0, r + 2)$ . Careful here! It is $r+2$ and not just $2$ . I spent more than 10 minutes wondering why I was getting stupid expressions while using just $2$ . :D

Now, let the point to be found make an angle $\alpha$ with the $y$ axis. So, its position is $(r\sin\alpha, r\cos\alpha)$ . The tangent to the circle at this point is given by the equation $x\sin\alpha + y\cos\alpha = r$ . This line passes through $(0, r+2)$ . Substitute that point in the equation and get the value of $\alpha$ by some algebraic manipulations. Once you find $\alpha$ , the required arc length (the final answer) is given as $r\cdot\alpha$ if $\alpha$ is in radians. Plug in the values and we're done!

Consider $\triangle OAB$ , where $O$ is the centre of the earth, $A$ is your far-away point, and $B$ is the top of the tower.

This is a right triangle with longest side $r+h$ and other sides $d$ and $r$ , where $h$ is the height of the tower and $d$ the distance we're interested in : $(r+h)^2 = r^2 + d^2$ so $d = \sqrt{2rh + h^2}$

Plugging in $r=6370$ and $h=2$ gives $d=159.637... \mbox{km}$

The problem becomes radically simple when sketched. Note that we form a right triangle with the highest point of the tower, $P$ , the center of the circle $O$ , and the segment connecting $P$ and any tangent point on the circle. By drawing the right triangle, we get that, for some radius $R$ and tower height $M$ : $\cos\theta = \frac{R}{R+M}$ Or, alternatively: $\theta = \cos^{-1}\frac{R}{R+M}$ Hence our curve now becomes: $R\theta = R\cdot\cos^{-1}\frac{R}{R+M}=6370\cdot\cos^{-1}\frac{6370}{6372}\approx 159.6\text{ km}$ And we are done.

Let $P$ be the farthest point from the tower which can be seen from the tower. It is easy to see that $P$ must be the point where the tangent from the top of the tower touches the surface of the Earth. To prove this, assume that there exists another point which is located further from the point of tangency and can be seen from the tower. Then, the straight line joining the top of the tower to that point must intersect the surface at a point located closer to it. This implies light cannot go to that point, as it cannot go through the surface of the Earth.

Let $O$ be the center of the Earth, and let $T$ be the tallest point on the tower. From our previous observation, $OP \perp TP$ . Let $\theta$ be the angle $\angle POT$ . Note that $\cos \theta= \frac{OP}{OT}= \frac{OP}{OP+PT}$ .

The length of the arc will then be $Radius \ of \ the \ Earth * \theta = 6370 \ \cos^{-1} \frac{6370}{6372} \ km \approx 159.603 \ km$ .