Angle in Biot-Savart formal in conductor loop I am referring to this problem in Griffiths:


My question: where does cos(theta) come from? r and dl are perpendicular to each other so with the formal from Biot-Savart, dlxr (vectors) = dl and nothing more. I see the explanation in parentheses but it still doesn't make sence. 
Any insight is appreciated. 
 A: Look at a single $dB$. Now compare it to a $dB$ contributed by an element of current on the opposite side of the ring. You will see their components that lie on the xy plane point opposite directions and thus cancel each other leaving you with just the $z$ component of each $dB$. Going around the full circle does this to each "$dB$" contributed by a $dl$ length of current. $\cos{\theta}$ will give you the adjacent side of the $dB$ right triangle in the diagram which points along the z axis - this is the $z$ component of $dB$. $\theta$ is the same for all points along the ring so $\cos{\theta}$ is a constant that can be moved outside the integral.
Also, $\vec{dl} \times \vec{r} \neq \vec{dl}$. It equals $dl$ multiplied by $r$ (where the sine of the angle between $dl$ and $r$ is 90 degrees and so sine of 90 degrees is just $1$). $r$ is the same for all points on the ring (all $dl$'s are the same distance away from the point in question) so $r$ can be taken outside the integral and you are left with just integrating $dl$ around the circle which is just the circumference of the circle.
