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From: Fred Nurk on 7 Jul 2010 05:46
VABC is a regular tetrahedron with base triangle ABC. (All faces are
equilateral triangles.) Find the magnitude of the angle between a sloping
edge and the base.

Let x be the length of any edge.
Why doesn't acos{[(x ^ 2 - (x / 2) ^ 2] ^ (1 / 2) / x} yield the right
answer?
From: Alois Steindl on 7 Jul 2010 05:51
Fred Nurk <albert.xtheunknown0(a)gmail.com> writes:

> VABC is a regular tetrahedron with base triangle ABC. (All faces are
> equilateral triangles.) Find the magnitude of the angle between a sloping
> edge and the base.
>
> Let x be the length of any edge.
> Why doesn't acos{[(x ^ 2 - (x / 2) ^ 2] ^ (1 / 2) / x} yield the right
> answer?
What makes you think it should?

Alois
From: Fred Nurk on 7 Jul 2010 06:09
Alois Steindl wrote:
> <snip>
> What makes you think it should?
> <snip>

The expression under the radical represents the length from a vertex on
the base triangle to the point half way of the opposite edge. x is the
length of a sloping edge. Then the inverse cosine of the quotient of the
expression and x should yield the correct answer, in my opinion.

Fred
From: Alois Steindl on 7 Jul 2010 06:14
Fred Nurk <albert.xtheunknown0(a)gmail.com> writes:

> Alois Steindl wrote:
>> <snip>
>> What makes you think it should?
>> <snip>
>
> The expression under the radical represents the length from a vertex on
> the base triangle to the point half way of the opposite edge. x is the
> length of a sloping edge. Then the inverse cosine of the quotient of the
> expression and x should yield the correct answer, in my opinion.
>
> Fred
If you look at the Tetrahedron from above, where do you think that you
see the top vertex?
Alois
From: Fred Nurk on 7 Jul 2010 06:32
Alois Steindl wrote:
> <snip>
> If you look at the Tetrahedron from above, where do you think that you
> see the top vertex?
> <snip>

At the centre of the base triangle.
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