This entry VI participated in the Math Carnival is hosted Sangakoo Blog. Take a long time without writing here and, so to speak, the size of this input is proportional to the time that I have not written one. But I hope that pleases from which to read.
* Sorry for the quality of the drawings, but most I can do. Suppose a
$ angle $ \\ alpha $ $ (always considered acute) formed by a straight $ $ r $ $ and a plane $ $ \\ pi $ $ . Suppose there is an observer to view $ $ O \\ in {\\ pi} $ $. What angle is a function of the original and the separation angle from the perpendicular to the line in the plane? Clearly, any observer at the perpendicular angle will be the same $ $ \\ alpha $ $, which, incidentally, is the least of all that can be seen in different positions around the cutoff line-plane. Also, an observer on the orthogonal projection of the line $ $ r $ $ $ at the $ \\ pi $ $, will ever see an angle $ $ \\ $ $ straight alpha_v whatever $ $ \\ alpha $ $. Also, an original right angle always look alike.
To arrive at an equation that proves the angle from the angle visible real $ $ \\ alpha $ $ and separation angle of the observation point on the perpendicular to the line in the plane , $ $ \\ beta $ $, I turned the system dihedral and trigonometry .
straight will use the maximum slope and maximum slope represented a dihedral plane. Example of maximum slope line:
The lines of maximum slope and maximum slope of a map indicating the tilt angle $ $ \\ alpha ^ {\\ prime} $ of the plane with the horizontal and vertical, respectively. Let's stay for example with the maximum slope.
Aware dimensional ( a little help for this ), the angle between the plane $ $ \\ delta $ $ with the horizontal is $ $ \\ alpha ^ {\\ prime} = tan ^ {-1} \\ frac {y} ^ {r_1} $ $. The triangle above the ground get:
$ $ y = m \\ cdot {} as $ $ $ $ \\ alpha $ $
Now we have to put $ $ r_1 $ $ on $ m $ function $ $. To do this, split up the triangle formed in the lower to the ground line.
$ $ l ^ 2 + r_1 ^ 2 = m ^ 2 $ $
$ $ h ^ 2 + (x) ^ 2 = l ^ 2 $ $
$ $ h ^ 2 + (x) ^ 2 + r_1 ^ 2 = m ^ 2 $ $
$ $ h ^ 2 + m ^ 2-2MX + x ^ 2 + r_1 ^ 2 = m ^ 2 $ $
$ $ h ^ 2 + x ^ 2 + r_1 ^ 2 = 2MX $ $ $ $ 2r_1
^ 2 = 2MX
$ $ $ $ r_1 = \\ sqrt [2] {x} $ $
$ $ h ^ 2 + (x) ^ 2 = l ^ 2 $ $
$ $ h ^ 2 + (x) ^ 2 + r_1 ^ 2 = m ^ 2 $ $
$ $ h ^ 2 + m ^ 2-2MX + x ^ 2 + r_1 ^ 2 = m ^ 2 $ $
$ $ h ^ 2 + x ^ 2 + r_1 ^ 2 = 2MX $ $ $ $ 2r_1
^ 2 = 2MX
$ $ $ $ r_1 = \\ sqrt [2] {x} $ $
Here it is noteworthy that in any right triangle is true:
$ $ x = m \\ cdot {} sin ^ 2 \\ gamma $ $
And continuing this:
$ $ r_1 = \\ sqrt [2] {m ^ 2 \\ cdot {} sin ^ 2 \\ gamma} $ $
$ $ r_1 = m \\ cdot { } sin \\ gamma $ $
$ $ r_1 = m \\ cdot { } sin \\ gamma $ $
Substituting $ $ and $ r_1 $ and $ $ $ $ in the first equation:
$ $ \\ alpha ^ {\\ prime} = tan ^ {-1} \\ frac {so \\ alpha} {sin \\ gamma} $ $
This equation determines the angle of the plane on the horizontal and vertical, as interpreted $ $ \\ Alpha $ $ and $ $ \\ gamma $ $.
To return to the original problem in dihedral, $ $ \\ alpha ^ {\\ prime} $ $ is the visible angle ($ $ \\ alpha_v $ $) of $ $ \\ alpha $ $ from the trace horizontal plane $ $ \\ delta_1 $ $ (intersection of horizontal line and the plane), in the case of the maximum slope line, or whatever it is, a $ $ (\\ frac {\\ pi} {2} - \\ gamma) $ of the perpendicular. Thus, $ $ \\ beta = \\ frac {\\ pi} {2} - \\ gamma $ $ and therefore $ $ \\ gamma = \\ frac {\\ pi} {2} - \\ beta $ $.
In a nutshell, that $ $ \\ alpha ^ {\\ prime} = \\ alpha_v $ $ taking into account that $ $ \\ beta = \\ frac {\\ pi} {2} - \\ gamma $ $.
$ $ \\ alpha ^ {\\ prime} $ $ is visible $ $ \\ alpha $ $ on a separate point of the perpendicular $ $ \\ frac {\\ pi} {2} - \\ gamma $ $.
Therefore, the equation that determines the angle visible from the same plane, $ $ \\ alpha_v $ $, depending on the original $ angle $ \\ alpha $ $ (formed by a line and a plane) and the separation angle the observer with respect to the perpendicular to the line in the plane, $ $ \\ beta $ $ is:
$ $ \\ alpha_v = tan ^ {-1} \\ frac {so \\ alpha} {cos \\ beta} $ $
From here, without further support a deduction, I concluded, despite the redundancy, how to vary the above equation for observation points outside the same plane ($ $ O \\ notin \\ pi $ $). I assumed that the variation of angle visible to the observation point move vertically around the cutoff of the line with the plan would occur in reverse order to that which occurs with the horizontal motion in the same plane around the cutoff point (with $ $ O \\ in {\\ pi }$$).
Thus, the resulting equation for vertical motion only, with the orthogonal projection of the observation point belongs always to the line in the plane perpendicular to that observed, would the following:
$ $ \\ alpha_ {v2} = tan ^ {-1} (tan \\ alpha \\ cdot {} cos \\ beta_V) $ $
And the combination of the two equations results in:
$ $ \\ alpha_ {visible} = tan ^ {-1} (tan \\ alpha \\ frac {cos \\ beta_V} {cos \\ beta_H}) $ $
Where $ $ \\ beta_V $ $ for the separation angle (vertical) point comments on the plane and centered at the cutoff line-plane, and $ $ \\ beta_H $ $ (the $ $ \\ beta $ $ to dry in the first part), the separation angle (horizontal) of the orthogonal projection by $ $ O $ $ on the plane perpendicular to the original line $ $ $ $ and r center at the same point of cut.
Thanks for reading. Everything is open to any corrections or suggestions.
