Find Angles of Intersecting Planes & Triangles - Stereometry Problem

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SUMMARY

This discussion focuses on solving a stereometry problem involving three intersecting planes that form a pyramid. The angles between the edges at the apex are given as ADB = 60°, ACD = 45°, and BCD = 45°. The user successfully determines the angle GHF using basic trigonometric functions (sine, cosine, tangent) without relying on complex formulas. The calculated angle GHF is 60°, which aligns with the user's expectation despite a conflicting reference from a textbook suggesting a different expression for the angle.

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  • Familiarity with geometric concepts of angles and triangles
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  • Study the properties of intersecting planes in three-dimensional geometry
  • Learn advanced trigonometric identities and their applications in geometry
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  • Investigate the derivation of angles in stereometric problems using formulas
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Students and educators in geometry, particularly those tackling stereometry problems, as well as anyone interested in understanding the practical application of trigonometric functions in three-dimensional shapes.

kernelpenguin
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I have 3 planes intersecting and forming a kind of a pyramid. The angles between pyramid's edges from the top point are known. How can I find the angles between side triangles? (Angles between each two planes from their intersection lines)

Note: Don't want formulaes here. Just understanding.
 
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Understand this, you need formulas to find the angles.
 
Well, I think I solved it. Without using any formulaes except for simple sin, cos and tan

http://hot.ee/kernelpenguin/mathsolution.png

Angles given: ADB = 60, ACD = 45, BCD = 45.
Need angle: GHF.
Assume: DH = 10. So this solution is a special-case scenario and won't get
full points on a test. However, the teacher demonstrated the "proper"
solution today and that took about 30 minutes and the whole blackboard so
this is considerably smaller.

1. Solve triangle GHD.
DH = 10
GDH = 45
GHD = 90 because GH is perpendicular to DB.
Need only GH and that is 10 after a quick tan()

2. Solve triangle FHD.
DH = 10
FDH = 30 because ED splits the corner at ADB.
FHD = 90 because FH is perpendicular to DB.
FH = 5 after a quick sin()

3. Solve triangle GFH.
GFH = 90
FH = 5 (from p2)
GH = 10 (from p1)
GHF = 60 after a quick cos()

Since there's bilateral symmetry in this "pyramid", the other such edge will
also be 60 degrees on the other side. Haven't even tried to calculate the top
one yet.

The weird thing is that the book says that the answer is arc cos sqrt(3)/3.
Which could in some weird ass way be 60 degrees. Or which could be wrong. Or
I could be wrong. No idea.

If you have time, see if my solution works or not :)
 
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