Attempting to figure out angle from ray of a cone in $\mathbb{R}^{3}$

Question

I'm working through Shifrin's Multivariable Mathematics and I've stumbled in attempting to figure out a specific statement. I have the following mapping:

Now one of the claims is that "if we fix $v = v_{0}$, the image is a ray making an angle of $\frac{\pi}{4}$." The claim of the image being a ray makes sense to me, but I was attempting to figure out how $\frac{\pi}{4}$ was arrived at. I tried to look at the cone through the point of view of an individual slice and use the trig functions available for right triangles with my $Z$ - axis serving as one of the two axis, but any manipulation in that manner doesn't get me the result.

I guess I'm wondering if it is possible to figure out the angle without having to resort to spherical coordinates?

Answer 1

Here is a diagram.

$r \leq \sqrt{u^2 \cos^2\theta + u^2 \sin^2\theta} = u$

So what you have is a right angled triangle with $\, l^2 = r^2 + z^2 = 2u^2 \implies l = u \sqrt2$.

Hence in this case, the angle between the axis of the cone and its outer edge is

$\displaystyle \tan^{-1} ({\frac{r}{z}}) = \tan^{-1} (1) = \frac{\pi}{4}$.