# Thread: How to find the altitude and size of an object atop a building

1. Originally Posted by stapel
One can find the height of the top of the antenna but, no, nothing else can be determined.
the height of the top of the antenna = 1.06233743 miles !!!!

2. Originally Posted by stapel
One can find the height of the top of the antenna but, no, nothing else can be determined.
Darn English, I should have said length of the antenna.

3. Originally Posted by Subhotosh Khan
the height of the top of the antenna = 1.06233743 miles !!!!
I'm getting something similar. For clarity, I'm using the following derived diagram:

Code:
diagram:
*
/|\
/ | \
/  h|  \
/    |   \
/      |    \
/60     |   70\
*--------*------*
|<--y--->|<---->|
5280-y
. . . . .$\dfrac{h}{y}\, =\, \tan(60^{\circ})$

. . . . .$\dfrac{h}{5,280\, -\, y}\, =\, \tan(70^{\circ})$

Solving for "y=":

. . . . .$\dfrac{h}{\tan(60^{\circ})}\, =\, y$

. . . . .$\dfrac{h}{\tan(70^{\circ})}\, =\, 5,280\, -\, y$

. . . . .$\dfrac{h}{\tan(70^{\circ})}\, -\,5,280\, =\, -y$

. . . . .$y\, =\, 5,280\, -\, \dfrac{h}{\tan(70^{\circ})}$

This allows me to form an equation only in "h", which is the only variable I care about anyway:

. . . . .$5,280\, -\, \dfrac{h}{\tan(70^{\circ})}\, =\, \dfrac{h}{\tan(60^{\circ})}$

. . . . .$5,280\, =\, h\, \left(\dfrac{1}{\tan(60^{\circ})}\, +\, \dfrac{1}{\tan(70^{\circ})}\right)$

. . . . .$\dfrac{5,280}{\left(\dfrac{1}{\tan(60^{\circ})}\, +\, \dfrac{1}{\tan(70^{\circ})}\right)}\,=\, h$

So the result is an utterly unreasonable value.

4. Using Law of Sines once, then the 30-60-90 triangle,
directly makes it 1.062337425275....as per Subhotosh...

k = SIN(70) / SIN(50)

h = k/2 * SQRT(3)