Rolling a cone on the $xy$ plane

Consider the cone in the initial position. We'll find the unit vector $\mathbf{a}$ along its axis. Since the semi-vertical angle is $\theta_c = 15^{\circ}$ , and the cone surface is tangent to the $xy$ plane, then the axis makes an angle of $\theta_c$ with the $xy$ plane, and similarly it makes an angle of $\theta_c$ with the $xz$ plane. If $\mathbf{k}$ is the unit vector along the $z$ -axis, then it follows that,

$\mathbf{a} \cdot \mathbf{k} = \cos( \dfrac{\pi}{2} - \theta_c ) = \sin \theta_c$

And similarly, if $\mathbf{j}$ is the unit vector along the $y$ -axis, then

$\mathbf{a} \cdot \mathbf{j} = \cos( \dfrac{\pi}{2} - \theta_c ) = \sin \theta_c$

If the unit vector $\mathbf{a} = (a_x, a_y, a_z )$ , then from the above equations,

$a_z = a_y = \sin \theta_c$

And, since $\mathbf{a}$ is a unit vector, then $a_x = \sqrt{1 - a_y^2 - a_z^2} = \sqrt{ 1 - 2 \sin^2 \theta_c } = \sqrt{ \cos 2 \theta_c }$

The line segment of tangency between the cone and the $xy$ plane has a length of the slant height of the cone, which is $s = \dfrac{h}{\cos \theta_c } = \dfrac{15}{\cos \theta_c }$ and points along the projection of the cone's axis onto the $xy$ plane, that is, along the vector $( \sqrt{ \cos 2 \theta_c} , \sin \theta_c )$ , thus the initial angle that this tangency line segment makes with the $x$ -axis is

$\phi_0 = \tan^{-1} \dfrac{ \sin \theta_c }{\sqrt{ \cos 2 \theta_c}}$

Now from the symmetry of the problem, it follows that the final angle (with the $x$ -axis ) is $\dfrac{\pi}{2} - \phi_0$ , Hence, the angle swept by the tangent line segment is $\Delta \phi = \dfrac{\pi}{2} - 2 \phi_0$ . So, finally, the area of the sector is

$\text{Sector Area} = \frac{1}{2} s^2 \Delta \phi$

And this evaluates to $\boxed{123.9857}$