# Thread: angle which is made out of one of the pyramids' bases and the base OABC?

1. ## angle which is made out of one of the pyramids' bases and the base OABC?

The figure shows a pyramid OABCD in a coordinate system with the starting point O. Determine the angle which is made out of one of the pyramids' bases and the base OABC.

I hope this is understandable, my english isn't really that good, due to teh fact that i'm from Denmark

The points are A(4,0,0), B(4,4,0), C(0,4,0), D(2,2,10) and O(0,0,0)

2. Originally Posted by Darkplayer
The figure shows a pyramid OABCD in a coordinate system with the starting point O. Determine the angle which is made out of one of the pyramids' bases and the base OABC.

I hope this is understandable, my english isn't really that good, due to teh fact that i'm from Denmark

The points are A(4,0,0), B(4,4,0), C(0,4,0), D(2,2,10) and O(0,0,0)

I've attached a rough sketch of the pyramid. The angle you are looking for is coloured in blue.

To determine the value of the angle use the indicated right triangle.

3. The situation looks like this:

4. Originally Posted by Darkplayer
The situation looks like this:

...
It looks pretty much the same as my sketch ...

Let $\displaystyle \alpha$ denote the angle between the base square and one side area of the pyramid. Then $\displaystyle \alpha$ can be calculated by:

$\displaystyle \tan(\alpha)=\dfrac{10}2~\implies~|\alpha| \approx 78.69^\circ$

5. You are kidding me right? Is it that simple?

6. Originally Posted by Darkplayer
You are kidding me right?
Onskyldt - but I'm never kidding anyone - ääähem ... mostly.

Is it that simple?
Yes. The pyramid is symmetric, the vertices have integer coordinates - so what did you expect?

(To prove my result you can use vectors:

The angle between 2 planes is as large as the angle between the normal vectors of the plane:

normal vector of the base: $\displaystyle \overrightarrow{n_{base}}=(0,0,1)$

normal vector of OCD: $\displaystyle \overrightarrow{n_{OCD}}=(0,1,0) \times (2,2,10) = (10,0,2)$

Then

$\displaystyle \cos(\alpha)=\dfrac{(0,0,1) \cdot (10,0,2)}{\sqrt{1} \cdot \sqrt{104}} = \dfrac2{\sqrt{104}}$

Now calculate $\displaystyle \alpha$ and you'll get exactly the same value as before.