accelerationkinematicsrotational-kinematicsvectorsvelocity
The book I am reading shows a proof of centripetal acceleration.
It proceeds to say that the linear velocity is always at a tangent to the radius, so the angle between $V_A$ and $V_B$ is also $\theta$
This is what it shows next. 
I can't seem to understand why the angle $\theta$ is the same.
Can someone explain please why, because I don't understand the book explanation? Thanks.
The relative velocity between two points depends on the frame.
In general, $A$ will see $B$ going at a different velocity than $B$ sees $A$.
Imagine that, in your example, $\mathbf{V_B}=\mathbf{V_A}$ (zero relative velocity) in the inertial frame. In the rotating frame, $B$ will lag a bit behind, because it is further than $A$ from the rotation center. This shows that the relative velocity is not zero in that frame.
Another example: Your friend is immobile a few meters away, while you are rotating around yourself. He will see you immobile, but you will see him turn around you. The relative velocities differ.
You should solve your problem in a given reference frame of your choice, and use the relative velocity in that frame.
The centripetal acceleration doesn't cause the tangential velocity, it has to be there to begin with in order for the body to orbit, as opposed to just fall to the surface. A good intuitive way of thinking about this is Newton's cannonball. If you place a cannon on a mountain and shoot a cannonball in the tangential direction, if the tangential velocity is small relative to the orbital velocity the cannon will fall to the surface. But as the tangential velocity gets larger, the cannon is still falling towards the surface but the Earth's surface is curving out from under it, so the distance between it and the surface shrinks more slowly despite the fact that the radial acceleration is the same, and the initial radial velocity was zero when it was fired. You can then imagine a series of cases where the tangential velocity gets larger and the fraction of the Earth's circumference traversed by the cannonball before hitting the surface gets larger, and the rate at which it approaches the surface decreases, until you reach a limit case where the rate at which it approaches the surface goes to zero and the cannonball is in a circular orbit, as depicted below:
There's a java applet here that allows you to play around with different initial tangential velocities for the projectile. Note that if you check the box "show full orbit" you can see what would happen if all the Earth's mass were concentrated at the center of the planet, so a cannonball fired from the same radius and with the same tangential velocity would be in some sort of elliptical orbit no matter how small the tangential velocity, an orbit that that can pass below the radius of the real Earth's surface (so in real life it would simply crash into the surface without making a full orbit). For a non-java illustration of this see the diagram in slide 23 here, attributed to Physics for the Enquiring Mind, Ch. 22, Fig. 29:
From the fact that all arcs of the cannonball can be understood as sections of possible elliptical orbits, you can deduce that even before the velocity needed for circular orbit, there would be a velocity that would result in an elliptical orbit that gets very close to the radius of the surface on the opposite side of the Earth from the cannon (ignoring atmospheric drag), then begins to climb in height again as it continues around the Earth, until returning to the height of the cannon at the position the cannon originally fired it.
Best Answer
The angle between the two velocity vectors is $x + x'$, where $x$ and $x'$ are the angles of the vectors with respect to a chord (the dashed line) of the circle. Now, as $ x + \alpha = \pi/2$ we have that $x = \pi/2 - \alpha$.
Similarly, $x' + \pi/2 + \alpha = \pi \Rightarrow x' = \pi/2 - \alpha$.
Thus, $$x + x' = \left(\frac{\pi}{2} - \alpha \right) + \left(\frac{\pi}{2} - \alpha \right) = \pi -2\alpha.$$
$\alpha$ is the angle appearing in the isosceles triangle so $2\alpha = \pi - \theta$ and therefore $$x+x' = \pi - (\pi - \theta) = \theta,$$ as required.