I'm getting something similar. For clarity, I'm using the following derived diagram:
. . . . .[tex]\dfrac{h}{y}\, =\, \tan(60^{\circ})[/tex]Code:diagram: * /|\ / | \ / h| \ / | \ / | \ /60 | 70\ *--------*------* |<--y--->|<---->| 5280-y
. . . . .[tex]\dfrac{h}{5,280\, -\, y}\, =\, \tan(70^{\circ})[/tex]
Solving for "y=":
. . . . .[tex]\dfrac{h}{\tan(60^{\circ})}\, =\, y[/tex]
. . . . .[tex]\dfrac{h}{\tan(70^{\circ})}\, =\, 5,280\, -\, y[/tex]
. . . . .[tex]\dfrac{h}{\tan(70^{\circ})}\, -\,5,280\, =\, -y[/tex]
. . . . .[tex]y\, =\, 5,280\, -\, \dfrac{h}{\tan(70^{\circ})}[/tex]
This allows me to form an equation only in "h", which is the only variable I care about anyway:
. . . . .[tex]5,280\, -\, \dfrac{h}{\tan(70^{\circ})}\, =\, \dfrac{h}{\tan(60^{\circ})}[/tex]
. . . . .[tex]5,280\, =\, h\, \left(\dfrac{1}{\tan(60^{\circ})}\, +\, \dfrac{1}{\tan(70^{\circ})}\right)[/tex]
. . . . .[tex]\dfrac{5,280}{\left(\dfrac{1}{\tan(60^{\circ})}\, +\, \dfrac{1}{\tan(70^{\circ})}\right)}\,=\, h[/tex]
So the result is an utterly unreasonable value.
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)
Last edited by Denis; 06-30-2016 at 09:22 PM.
I'm just an imagination of your figment !
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