Everyone knows that the solid angle subtended at the corner of a cube
is equal to pi/2. Can you work out the analytical value of the solid angle subtended at the corner of
(a) a regular tetrahedron,
and (b) a regular octagon?
Hint: You cant solve (a) without solving (b).

Originally posted by rspoddar82 Everyone knows that the solid angle subtended at the corner of a cube
is equal to pi/2. Can you work out the analytical value of the solid angle subtended at the corner of
(a) a regular tetrahedron,
and (b) a regular octagon?
Hint: You cant solve (a) without solving (b).

The answer to b) is 0, but I don't see what this has to do with a) ðŸ˜›

Originally posted by Acolyte The answer to b) is 0, but I don't see what this has to do with a) ðŸ˜›

actually (b) should read "a regular octahedron" . My error . But without finding the solid angle at the corner of a regular octahedron U can't find the solid angle at the corner of a regular ettrahedronmi

yes acolyte U have Got it. The answer is indeed
(a)3sec^-1(3) - pi which is equal to 3cos^-1(1/3) - pi.
But how could U solve it without solving for the solid angle at the corner of a regular octahedron?

Originally posted by rspoddar82 yes acolyte U have Got it. The answer is indeed
(a)3sec^-1(3) - pi which is equal to 3cos^-1(1/3) - pi.
But how could U solve it without solving for the solid angle at the corner of a regular octahedron?

acolyte has found the value of the solid angle at the corner of a regular tetrahedron. But how is it related to the solid angle at the corner of a regular octehedron? AND WHAT IS ITS VALUE?

This brings me to a funny story. My chemistry teacher told us some things about molecules and then that the bond angle in tetrahedral molecules was around 109.5 degrees. I told him:

There are five atoms in a tetrahedral molecule, at positions P1, P2, P3, P4, P5. Say P3 is the 'central' one. The vector P1P3 we'll call u, P2P3 v, P4P3 w, P5P3 x. The bond length is l = |u| = |v| = |w| = |x|. So, since the molecule is such that every interatomic distance is maximized relative to the others, any pair of atoms could be interchanged (rotation about P3) so that:

u + v + w + x = 0

Thus u*(u + v + w + x) = 0 (dot product here). This means:

l^2 + 3l^2 cos A = 0

where A is the bond angle (P3 to any of the others). Dividing by l^2 gives cos A = -1/3, so A is about 109.47 degrees (since obviously a bond angle is between 0 and 180 degrees).

The funny bit is that I very nearly failed chemistry.

(Acolyte's result can be got by taking a cross-section through any outer point and P3 and looking at one of the isoceles triangles, but his way doesn't involve the above rigamarole.)

Originally posted by royalchicken This brings me to a funny story. My chemistry teacher told us some things about molecules and then that the bond angle in tetrahedral molecules was around 109.5 degrees. I told him:

There are five atoms in a tetrahedral molecule, ...[text shortened]... es (since obviously a bond angle is between 0 and 180 degrees).

Did your chemistry teacher say, "Pardon??"

Originally posted by royalchicken This brings me to a funny story. My chemistry teacher told us some things about molecules and then that the bond angle in tetrahedral molecules was around 109.5 degrees. I told him:

There are five atoms in a tetrahedral molecule, at positions P1, P2, P3, P4, P5. Say P3 is the 'central' one. The vector P1P3 we'll call u, P2P3 v, P4P3 w, P5P3 x. ...[text shortened]... nd looking at one of the isoceles triangles, but his way doesn't involve the above rigamarole.)

the story of ur chemistry teacher is interesting. by the way can ur method be applied in exactly similar way to find out the solid angle at the corner of a regular octahedron?/

Originally posted by cheskmate the story of ur chemistry teacher is interesting. by the way can ur method be applied in exactly similar way to find out the solid angle at the corner of a regular octahedron?/

Will this be solved , by finding out the area of a spherical square(area on the surface of a sphere of unit radius) of side length equal to the magnitude of the plane angle pi/3 ?? This is the method used by one of the respondents above..He has found the area of a spherical equilateral triangle of side length = pi/3. Intutively , by finding the area of a spherical square of the same side length should give the solid angle of the octahedron.
Can anybody say something definite on this and enlighten whether my intution is on the right track?

Originally posted by ranjan sinha Will this be solved , by finding out the area of a spherical square(area on the surface of a sphere of unit radius) of side length equal to the magnitude of the plane angle pi/3 ?? This is the method used by one of the respondents above..He has found the area of a spherical equilateral triangle of side length = pi/3. Intutively , by finding the area ...[text shortened]... anybody say something definite on this and enlighten whether my intution is on the right track?

No, Acolyte's method will not work in this case.....

Originally posted by cheskmate acolyte has found the value of the solid angle at the corner of a regular tetrahedron. But how is it related to the solid angle at the corner of a regular octehedron? AND WHAT IS ITS VALUE?

well ,I guess the value of solid angle at the corner of an octahedron is pie/12..