# Intersection between 2 planes

I need help understanding the formula for intersection between 2 planes

Where is $\alpha$ actually, its not clear in the diagram. It appears that I could just interpret it as the below?

Where $\beta = \theta$?

UPDATE

Now given the question:

What I did

Whats wrong? Correct answer $60.7^{\circ}$. Did I "expanded" the equation for plane whose equation was in Cartesian form correctly?

-
The diagram is somewhat confusing, but: remember that two sets of angles are formed when two non-perpendicular lines cross. You have one set of obtuse "vertical angles", and you have one set of acute "vertical angles". The acute and obtuse angles are complementary; that is, letting $\theta$ be the acute angle and $\alpha$ be the obtuse one, $\alpha+\theta=\pi$. –  Ｊ. Ｍ. Nov 14 '11 at 2:46

Maybe it's best to ignore the diagram from your book. Just consider two planes with normal vectors ${\bf n}_1$ and ${\bf n}_2$. Let's suppose that the angle between these vectors is $\alpha$ (which is acute). Then that's the angle between the two planes.
Now what if our first plane had had the normal vector $-{\bf n}_1$? Then the angle between our two normals would have been $\theta$. This angle is obtuse. So it should be replaced by $\alpha = \pi - \theta$.