# angle between faces of a pyramid

I have a square based pyramid with vertices at $O(0,0,0)$ $A(1,0,0)$, $B(1,1,0)$, $C(0,1,0)$, and $D(0,0,1)$.

I need to find the angle between the faces $ABD$ and $BCD$ $(1,0,1)$ and $(0,1,1)$.

When I use the angle formula I get a solution of $\theta =60°$.

However the solution says that it should be $120°$ why is the obtuse angle correct in this case?

• Make a sketch of the pyramid and choose the direction of the vectors accordingly. – user May 12 '18 at 17:42
• That depends on the definition of the angle between two faces. I would say that it is common to define the angle between two planes to be in the interval $[0,90^\circ]$. However, those faces are not full planes. In that case a natural definition would be the dihedral angle between the two half-planes bordered by the shared edge. That angle can easily be obtuse. – Jyrki Lahtonen May 12 '18 at 17:49
• Indeed the solution 60° seems more natural to me adopting the other direction for the normal vectors. – user May 12 '18 at 17:58

Note that it depends upon the orientation you have used for the vectors, indeed

$$60°=180°-120°$$

then your calculation is almost correct but you need to fix orientations according to the convention adopted.

If we choose as direction of the normal vectors the outer direction, your calculation is correct, notably we have

• normal to face $ADB$: $v=DA\times DB=(1,0,-1)\times(1,1,-1)=(1,0,1)$

• normal to face $BCD$: $w=DB\times DC=(1,1,-1)\times(1,1,-1)=(0,1,1)$

then

$$\cos \theta =\frac{v\cdot w}{|v||w|}=\frac{1}{2}\implies \theta=60°$$

As observed by Jyrki Lahtonen, if we consider the dihedral angle between the two half-planes bordered by the shared edge, it can be easily shown that this angle is exactly $180°-60°=120°$.