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 $\displaystyle p_{ABC}$ and $\displaystyle p_{AFC}$.
3. The angle included by the two planes equals the angle which is included by the two normal vectors.
4. Let $\displaystyle \phi$ denote the angle included by the normal vectors. Use $\displaystyle \cos(\phi)=\dfrac{\overrightarrow{n_{ABC}} \cdot \overrightarrow{n_{AFC}}}{|\overrightarrow{n_{ABC} }| \cdot |\overrightarrow{n_{AFC}}|}$
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?
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.
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 $\displaystyle \Delta(CAB)$ and $\displaystyle \Delta(CAF)$, q and s respectively.
Since $\displaystyle \Delta(CAB)$ is an isosceles right triangle the length of $\displaystyle q = \frac12 \sqrt{2}$. The line segments p, q and s form a right triangle. The angle between the triangles $\displaystyle \Delta(CAB)$ and $\displaystyle \Delta(CAF)$ is:
$\displaystyle \alpha = \arctan\left(\dfrac1{\frac12 \sqrt{2}} \right)\approx 54.7^\circ$