Angle Between Two Faces of a Triangular Pyramid
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A regular triangular pyramid has all of edges of length 36. What is the angle between any two faces of this pyramid?
I have no idea what to do for this problem. Can someone explain to me what an angle between two faces is?
geometry
asked Jan 16 at 1:19
semicolon822semicolon822
51
$endgroup$
add a comment |
$begingroup$
A regular triangular pyramid has all of edges of length 36. What is the angle between any two faces of this pyramid?
I have no idea what to do for this problem. Can someone explain to me what an angle between two faces is?
geometry
asked Jan 16 at 1:19
semicolon822semicolon822
51
$endgroup$
$begingroup$
Why do edge lengths have anything to do with this problem?
$endgroup$
– David G. Stork
Jan 16 at 1:24
$begingroup$
idk thats what the problem said
$endgroup$
– semicolon822
Jan 16 at 1:25
add a comment |
$begingroup$
A regular triangular pyramid has all of edges of length 36. What is the angle between any two faces of this pyramid?
I have no idea what to do for this problem. Can someone explain to me what an angle between two faces is?
geometry
asked Jan 16 at 1:19
semicolon822semicolon822
51
$endgroup$
A regular triangular pyramid has all of edges of length 36. What is the angle between any two faces of this pyramid?
I have no idea what to do for this problem. Can someone explain to me what an angle between two faces is?
geometry
geometry
asked Jan 16 at 1:19
semicolon822semicolon822
51
asked Jan 16 at 1:19
semicolon822semicolon822
51
asked Jan 16 at 1:19
semicolon822semicolon822
51
asked Jan 16 at 1:19
semicolon822semicolon822
51
asked Jan 16 at 1:19
semicolon822semicolon822
51
51
$begingroup$
Why do edge lengths have anything to do with this problem?
$endgroup$
– David G. Stork
Jan 16 at 1:24
$begingroup$
idk thats what the problem said
$endgroup$
– semicolon822
Jan 16 at 1:25
add a comment |
$begingroup$
Why do edge lengths have anything to do with this problem?
$endgroup$
– David G. Stork
Jan 16 at 1:24
$begingroup$
idk thats what the problem said
$endgroup$
– semicolon822
Jan 16 at 1:25
$begingroup$
Why do edge lengths have anything to do with this problem?
$endgroup$
– David G. Stork
Jan 16 at 1:24
$begingroup$
Why do edge lengths have anything to do with this problem?
$endgroup$
– David G. Stork
Jan 16 at 1:24
$begingroup$
idk thats what the problem said
$endgroup$
– semicolon822
Jan 16 at 1:25
$begingroup$
idk thats what the problem said
$endgroup$
– semicolon822
Jan 16 at 1:25
add a comment |
2 Answers
2
active
oldest
votes
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The lengths of the sides of the regular tetrahedron are irrelevant.
Hint: Find the vectors from the tetrahedron's center to the center of two faces. Then calculate the angle between them.
answered Jan 16 at 1:47
David G. StorkDavid G. Stork
11.6k41533
$endgroup$
$begingroup$
could you elaborate on that
$endgroup$
– semicolon822
Jan 16 at 1:49
$begingroup$
Could you show what work you've done? Where you've gotten stuck? What isn't clear about my posting? What specifically you don't understand?
$endgroup$
– David G. Stork
Jan 16 at 1:49
$begingroup$
i dont know how to "find the vectors"
$endgroup$
– semicolon822
Jan 16 at 1:50
$begingroup$
Vector: A line from the tetrahedron center (average of four vertices) to a face center (average of three vertices).
$endgroup$
– David G. Stork
Jan 16 at 1:54
$begingroup$
Further hint: a regular tetrahedron is vertex inscribable into a cube. Thus the face vectors of the tetrahedron just point to the omitted vertices of that cube.
$endgroup$
– Dr. Richard Klitzing
Jan 16 at 17:29
add a comment |
$begingroup$
Let the regular tetrahedron be ABCD with ABC as the base, D as the highest vertex, ADB and and CDB as its slant faces.
From ABD (which is an equilateral triangle), we need to find the altitude AX.
Similarly, find CX from CBD. Note that since the tetrahedron is regular, the above mentioned two altitudes will (1) meet at the same point X; and (2) have the same length.
The required angle can be found by applying cosine law to the triangle AXC.
answered Jan 16 at 4:46
MickMick
12k21641
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The lengths of the sides of the regular tetrahedron are irrelevant.
Hint: Find the vectors from the tetrahedron's center to the center of two faces. Then calculate the angle between them.
answered Jan 16 at 1:47
David G. StorkDavid G. Stork
11.6k41533
$endgroup$
$begingroup$
could you elaborate on that
$endgroup$
– semicolon822
Jan 16 at 1:49
$begingroup$
Could you show what work you've done? Where you've gotten stuck? What isn't clear about my posting? What specifically you don't understand?
$endgroup$
– David G. Stork
Jan 16 at 1:49
$begingroup$
i dont know how to "find the vectors"
$endgroup$
– semicolon822
Jan 16 at 1:50
$begingroup$
Vector: A line from the tetrahedron center (average of four vertices) to a face center (average of three vertices).
$endgroup$
– David G. Stork
Jan 16 at 1:54
$begingroup$
Further hint: a regular tetrahedron is vertex inscribable into a cube. Thus the face vectors of the tetrahedron just point to the omitted vertices of that cube.
$endgroup$
– Dr. Richard Klitzing
Jan 16 at 17:29
add a comment |
$begingroup$
The lengths of the sides of the regular tetrahedron are irrelevant.
Hint: Find the vectors from the tetrahedron's center to the center of two faces. Then calculate the angle between them.
answered Jan 16 at 1:47
David G. StorkDavid G. Stork
11.6k41533
$endgroup$
$begingroup$
could you elaborate on that
$endgroup$
– semicolon822
Jan 16 at 1:49
$begingroup$
Could you show what work you've done? Where you've gotten stuck? What isn't clear about my posting? What specifically you don't understand?
$endgroup$
– David G. Stork
Jan 16 at 1:49
$begingroup$
i dont know how to "find the vectors"
$endgroup$
– semicolon822
Jan 16 at 1:50
$begingroup$
Vector: A line from the tetrahedron center (average of four vertices) to a face center (average of three vertices).
$endgroup$
– David G. Stork
Jan 16 at 1:54
$begingroup$
Further hint: a regular tetrahedron is vertex inscribable into a cube. Thus the face vectors of the tetrahedron just point to the omitted vertices of that cube.
$endgroup$
– Dr. Richard Klitzing
Jan 16 at 17:29
add a comment |
$begingroup$
The lengths of the sides of the regular tetrahedron are irrelevant.
Hint: Find the vectors from the tetrahedron's center to the center of two faces. Then calculate the angle between them.
answered Jan 16 at 1:47
David G. StorkDavid G. Stork
11.6k41533
$endgroup$
The lengths of the sides of the regular tetrahedron are irrelevant.
Hint: Find the vectors from the tetrahedron's center to the center of two faces. Then calculate the angle between them.
answered Jan 16 at 1:47
David G. StorkDavid G. Stork
11.6k41533
answered Jan 16 at 1:47
David G. StorkDavid G. Stork
11.6k41533
answered Jan 16 at 1:47
David G. StorkDavid G. Stork
11.6k41533
answered Jan 16 at 1:47
David G. StorkDavid G. Stork
11.6k41533
11.6k41533
$begingroup$
could you elaborate on that
$endgroup$
– semicolon822
Jan 16 at 1:49
$begingroup$
Could you show what work you've done? Where you've gotten stuck? What isn't clear about my posting? What specifically you don't understand?
$endgroup$
– David G. Stork
Jan 16 at 1:49
$begingroup$
i dont know how to "find the vectors"
$endgroup$
– semicolon822
Jan 16 at 1:50
$begingroup$
Vector: A line from the tetrahedron center (average of four vertices) to a face center (average of three vertices).
$endgroup$
– David G. Stork
Jan 16 at 1:54
$begingroup$
Further hint: a regular tetrahedron is vertex inscribable into a cube. Thus the face vectors of the tetrahedron just point to the omitted vertices of that cube.
$endgroup$
– Dr. Richard Klitzing
Jan 16 at 17:29
add a comment |
$begingroup$
could you elaborate on that
$endgroup$
– semicolon822
Jan 16 at 1:49
$begingroup$
Could you show what work you've done? Where you've gotten stuck? What isn't clear about my posting? What specifically you don't understand?
$endgroup$
– David G. Stork
Jan 16 at 1:49
$begingroup$
i dont know how to "find the vectors"
$endgroup$
– semicolon822
Jan 16 at 1:50
$begingroup$
Vector: A line from the tetrahedron center (average of four vertices) to a face center (average of three vertices).
$endgroup$
– David G. Stork
Jan 16 at 1:54
$begingroup$
Further hint: a regular tetrahedron is vertex inscribable into a cube. Thus the face vectors of the tetrahedron just point to the omitted vertices of that cube.
$endgroup$
– Dr. Richard Klitzing
Jan 16 at 17:29
$begingroup$
could you elaborate on that
$endgroup$
– semicolon822
Jan 16 at 1:49
$begingroup$
could you elaborate on that
$endgroup$
– semicolon822
Jan 16 at 1:49
$begingroup$
Could you show what work you've done? Where you've gotten stuck? What isn't clear about my posting? What specifically you don't understand?
$endgroup$
– David G. Stork
Jan 16 at 1:49
$begingroup$
Could you show what work you've done? Where you've gotten stuck? What isn't clear about my posting? What specifically you don't understand?
$endgroup$
– David G. Stork
Jan 16 at 1:49
$begingroup$
i dont know how to "find the vectors"
$endgroup$
– semicolon822
Jan 16 at 1:50
$begingroup$
i dont know how to "find the vectors"
$endgroup$
– semicolon822
Jan 16 at 1:50
$begingroup$
Vector: A line from the tetrahedron center (average of four vertices) to a face center (average of three vertices).
$endgroup$
– David G. Stork
Jan 16 at 1:54
$begingroup$
Vector: A line from the tetrahedron center (average of four vertices) to a face center (average of three vertices).
$endgroup$
– David G. Stork
Jan 16 at 1:54
$begingroup$
Further hint: a regular tetrahedron is vertex inscribable into a cube. Thus the face vectors of the tetrahedron just point to the omitted vertices of that cube.
$endgroup$
– Dr. Richard Klitzing
Jan 16 at 17:29
$begingroup$
Further hint: a regular tetrahedron is vertex inscribable into a cube. Thus the face vectors of the tetrahedron just point to the omitted vertices of that cube.
$endgroup$
– Dr. Richard Klitzing
Jan 16 at 17:29
add a comment |
$begingroup$
Let the regular tetrahedron be ABCD with ABC as the base, D as the highest vertex, ADB and and CDB as its slant faces.
From ABD (which is an equilateral triangle), we need to find the altitude AX.
Similarly, find CX from CBD. Note that since the tetrahedron is regular, the above mentioned two altitudes will (1) meet at the same point X; and (2) have the same length.
The required angle can be found by applying cosine law to the triangle AXC.
answered Jan 16 at 4:46
MickMick
12k21641
$endgroup$
add a comment |
$begingroup$
Let the regular tetrahedron be ABCD with ABC as the base, D as the highest vertex, ADB and and CDB as its slant faces.
From ABD (which is an equilateral triangle), we need to find the altitude AX.
Similarly, find CX from CBD. Note that since the tetrahedron is regular, the above mentioned two altitudes will (1) meet at the same point X; and (2) have the same length.
The required angle can be found by applying cosine law to the triangle AXC.
answered Jan 16 at 4:46
MickMick
12k21641
$endgroup$
add a comment |
$begingroup$
Let the regular tetrahedron be ABCD with ABC as the base, D as the highest vertex, ADB and and CDB as its slant faces.
From ABD (which is an equilateral triangle), we need to find the altitude AX.
Similarly, find CX from CBD. Note that since the tetrahedron is regular, the above mentioned two altitudes will (1) meet at the same point X; and (2) have the same length.
The required angle can be found by applying cosine law to the triangle AXC.
answered Jan 16 at 4:46
MickMick
12k21641
$endgroup$
Let the regular tetrahedron be ABCD with ABC as the base, D as the highest vertex, ADB and and CDB as its slant faces.
From ABD (which is an equilateral triangle), we need to find the altitude AX.
Similarly, find CX from CBD. Note that since the tetrahedron is regular, the above mentioned two altitudes will (1) meet at the same point X; and (2) have the same length.
The required angle can be found by applying cosine law to the triangle AXC.
answered Jan 16 at 4:46
MickMick
12k21641
edited Jan 16 at 4:52
answered Jan 16 at 4:46
MickMick
12k21641
answered Jan 16 at 4:46
MickMick
12k21641
answered Jan 16 at 4:46
MickMick
12k21641
12k21641
add a comment |
add a comment |
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$begingroup$
Why do edge lengths have anything to do with this problem?
$endgroup$
– David G. Stork
Jan 16 at 1:24
$begingroup$
idk thats what the problem said
$endgroup$
– semicolon822
Jan 16 at 1:25