# Cube, Dihedral

• Mar 26th 2011, 03:02 PM
thamathkid1729
Cube, Dihedral
The figure below shows a cube
ABCDEFGH. Triangles ABC and AFC form an angle that is called dihedral because it is formed by two intersecting planes. Notice that the line of intersection is AC. Calculate the size of this angle, to the nearest tenth of a degree.

Attachment 21276

• Mar 26th 2011, 10:38 PM
earboth
Quote:

Originally Posted by thamathkid1729
[LEFT]The figure below shows a cube ABCDEFGH. Triangles ABC and AFC form an angle that is called dihedral because it is formed by two intersecting planes. Notice that the line of intersection is AC. Calculate the size of this angle, to the nearest tenth of a degree.

I can't help you because I don't know what you have done so far to solve this question.

Here are the steps which I would take:

1. Use coordinates and vectors: B(0, 0, 0), A(1, 0, 0), C(0, 1, 0) F(0, 0, 1)

2. Determine the normal vectors of the planes $p_{ABC}$ and $p_{AFC}$.

3. The angle included by the two planes equals the angle which is included by the two normal vectors.

4. Let $\phi$ denote the angle included by the normal vectors. Use $\cos(\phi)=\dfrac{\overrightarrow{n_{ABC}} \cdot \overrightarrow{n_{AFC}}}{|\overrightarrow{n_{ABC} }| \cdot |\overrightarrow{n_{AFC}}|}$
• Mar 27th 2011, 03:00 PM
thamathkid1729
My understanding of this problem is that I want to find the measure of angle BCF. Couldn't I just use the properties of a cube and trigonometry to determine that this angle is 45 degrees? Then why would it say to calculate to the nearest tenth of a degree? Or am I understanding this problem wrong?
• Mar 27th 2011, 08:51 PM
Soroban
Hello, thamathkid1729!

Angle FAB is 45 degrees; angle FCB is 45 degrees.

But the angle between the two triangles is not 45 degrees.

Look at it this way . . .

We have a cube (labeled ABCDEFGH).

Using a sharp knife, cut straight through vertices A, C, and F.
Remove the upper (larger) potion of the cube.
We will have a "corner" of the cube.

Code:

    F *       |* *       | *  *       |  *    *       |  *      *       |    *        *       |    *          * A       |      *        *       |      *    *       |        *  *     B *---------*                 C

We want the angle between the slanted face and the "floor".

earboth has the best plan for this.

• Mar 28th 2011, 02:40 AM
earboth
Quote:

Originally Posted by thamathkid1729
The figure below shows a cube
ABCDEFGH. Triangles ABC and AFC form an angle that is called dihedral because it is formed by two intersecting planes. Notice that the line of intersection is AC. Calculate the size of this angle, to the nearest tenth of a degree.

Attachment 21276

I've found a different way to determine the angle between the two planes.

The boundaries of an angle are straight lines, it's legs. If you want to determine the angle between two planes you have to look for two lines which include the same angle as the two planes. I use the heights in the triangles $\Delta(CAB)$ and $\Delta(CAF)$, q and s respectively.

Since $\Delta(CAB)$ is an isosceles right triangle the length of $q = \frac12 \sqrt{2}$. The line segments p, q and s form a right triangle. The angle between the triangles $\Delta(CAB)$ and $\Delta(CAF)$ is:

$\alpha = \arctan\left(\dfrac1{\frac12 \sqrt{2}} \right)\approx 54.7^\circ$