The angle of the sector not taken is $2\pi-\theta$. The length of the circular arc with central angle $2\pi-\theta$ radians and radius $r$ is $r(2\pi-\theta)=2\pi r-\theta r$ (by the definition of radian measure). When you form the cone, this will be the circumference of the circle at the base of the cone. The radius of that circle is $\frac{2\pi r-\theta r}{2\pi}=r\left(1-\frac{\theta}{2\pi}\right)$ (call it $R$).
The "slant height" of the cone is the radius of the original circle, $r$, and is the hypotenuse of a right triangle with base $R=r\left(1-\frac{\theta}{2\pi}\right)$ and height $h$ (let's say). Therefore
$$h=\sqrt{r^2-\left[ r\left(1-\frac{\theta}{2\pi}\right) \right]^2}$$
$$=r\sqrt{\frac{\theta}{\pi}-\frac{\theta^2}{4\pi^2}}$$
Therefore the volume of the cone will be
$$V=\frac 13\pi R^2h=\frac 13\pi\left[r\left(1-\frac{\theta}{2\pi}\right)\right]^2\left[r\sqrt{\frac{\theta}{\pi}-\frac{\theta^2}{4\pi^2}}\right]$$
$$=\frac 13\pi r^3\left(1-\frac{\theta}{2\pi}\right)^2\left(\sqrt{\frac{\theta}{\pi}-\frac{\theta^2}{4\pi^2}}\right)$$
You could simplify that further, as you like.
Note that the answer would have been significantly easier if you defined $\theta$ to be the central angle of the sector left after cutting rather than of the sector that was cut. If we define $\Theta$ as that central angle that was left, we end up with
$$V=\frac 13\pi r^3\left(\frac{\Theta}{2\pi}\right)^2
\sqrt{1-\left(\frac{\Theta}{2\pi}\right)^2}$$
The length of a circular arc is directly proportional to its central angle, so when you cut out a sector of angle $\theta$ and glue the cut edges together, the circumference of the base of the cone is going to be smaller than that of the original circle by this proportional amount: $c=C-\theta R$. Since the circumference of a circle is $2\pi$ times its radius, we then have for the radius of the cone’s base $$r = \left(1-\frac\theta{2\pi}\right)R.$$
The slant height of the cone is the original circle’s radius $R$, so $$\sin \alpha=\frac r R=1-\frac\theta{2\pi},$$ where $\alpha$ is the half-angle at the cone’s vertex.
If you cut out half of the circle, you’ll have $\sin\alpha=\frac12$, or $\alpha=\frac\pi6$, which is $30$ degrees.
Best Answer
The length of the edge of the sector is $2\pi r$ because that is the circumference of the base of the cone.
The radius of the sector is $l$ because that is the distance from the tip of the cone to the base, which becomes the distance from the center of the circle to the edge.
For any arc of length $L$ on a circle of radius $R$, the angle that it subtends is $\frac{L}{R}$ radians, essentially by definition.
Thus, the angle of the sector is $\theta=\frac{2\pi r}{l}$.