The lateral surface of a cone is twice the area of its base. Find the angle in the development of the lateral surface of the cone.

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Let $r$ be the radius of the base of the given cone, $l$ be the slant height, $S_{1}$ be the area of the base of the cone, and $S_{2}$ be the area of the lateral surface. Then $S_{2}$ is the area of the sector of a circle with radius $l$, which is the development of the lateral surface of the cone. According to the problem, $S_{2}=2 \cdot S_{1}$, or $\pi r l=2 \pi r^{2}$. From this, we find that $r=l / 2$. Therefore, the area of the sector is $\pi l^{2} / 2$, which is half the area of a circle with radius $l$. This means that the sector is a semicircle. Therefore, the angle in the development of the lateral surface of the cone is $180^{\circ}$.

## Answer

$180^{\circ}$.