# Thread: Triangles, Planes and 3D geometry

1. ## Triangles, Planes and 3D geometry

The Question

Given a rectangular tetrahedron (i.e. a pyramid where each face is an equilateral triangle), find the cosine of the angle between two of the faces.

My Workings

To be honest I have not really got any worth showing for this one, I seem to be struggling with the concept of questions that ask for the angles between 2 planes. I'm just not sure how to approach them.

Apparently cos(60) is not the answer

2. Originally Posted by AliceFisher
The Question

Given a rectangular tetrahedron (i.e. a pyramid where each face is an equilateral triangle), find the cosine of the angle between two of the faces.

My Workings

To be honest I have not really got any worth showing for this one, I seem to be struggling with the concept of questions that ask for the angles between 2 planes. I'm just not sure how to approach them.

Apparently cos(60) is not the answer
Hi Alice,

imagine you can stand all 3 sides of the tetrahedron perpendicular to the base,
as if the sides were only touching but can be pulled away from each other,
so that all 3 sides are now standing at right-angles to the base.

The vertical height "h" of any of these triangular faces in terms of the base edge length "x" is found using

$\displaystyle \frac{x}{2}=\frac{h}{tan60^o}$

$\displaystyle x=\frac{2h}{tan60^o}=\frac{2h}{\sqrt{3}}$

$\displaystyle h=\frac{\sqrt{3}}{2}x$

Now tilt the sides until they form the tetrahedron again,
and referring to the attachment,
if we have a second blue line going to the third apex of the base not shown in the attachment (think of the 3-D picture),
then we have the triangle that forms the angle between the faces.

It's dimensions are $\displaystyle x,\ \frac{\sqrt{3}}{2}x,\ \frac{\sqrt{3}}{2}x$

Then the cosine rule finds the angle between the faces

$\displaystyle x^2=\left(\frac{\sqrt{3}}{2}\right)^2x^2(2)-2\left(\frac{\sqrt{3}}{2}x\frac{\sqrt{3}}{2}x\righ t)cos\theta$

$\displaystyle x^2=\frac{6}{4}x^2-\frac{3}{2}x^2cos\theta$

$\displaystyle 3x^2cos\theta=3x^2-2x^2=x^2$

$\displaystyle 3cos\theta=1$

$\displaystyle \theta=cos^{-1}\left(\frac{1}{3}\right)=70.52878^o$

A 3-D picture would be more helpful here,
hence an oblique 3-D view is also attached,
where we need the angle between the blue lines identified by the arrows.
The angle at the base and at the apex is 60 degrees,
but the blue lines have a smaller length to the edges,
hence the angle between the faces is greater than 60 degrees.

Sorry there is a typo in the diagram.

The length of the blue line is $\displaystyle \frac{\sqrt{3}}{2}x$

3. Thanks for this I'm going to work through your answer now

4. Okay its been a bit of work but I think I have finally got it! Thanks again

The blue lines being longer in the middle really helped me understand why the angle would be different.

5. Hi Alice,

the attached drawing might be useful.
It shows that the angle between the pairs of blue lines are the same.
It's most convenient for us to calculate angle ABC, however.

The blue lines are on the base and the backface of the tetrahedron.