(by CHATGPT) We divide the sphere into four congruent spherical triangles by placing a triangular pyramid inside the sphere with its base coinciding with the given triangle. Let's call these four triangular pyramids $P_1, P_2, P_3,$ and $P_4$. First, let's prove that the sum of the angles of each of these triangular pyramids is less than $2\pi$. Since each side of the given triangle is an arc of a circle of radius $1$, the lengths of the sides are less than $2\pi$. Therefore, the angles of the given triangle are less than $\pi$. Now, consider one of the four pyramids, for example, $P_1$. The surface area of $P_1$ is equal to a quarter of the surface area of the sphere, so each face of $P_1$ has an area equal to one-fourth of the total surface area of the sphere. Since the surface area of a sphere is $4\pi$, each face of $P_1$ has an area of $\pi$. Now, we claim that each face of $P_1$ is congruent to the given triangle. Suppose not, and there exists a face of $P_1$ that is not congruent to the given triangle. Then, there must be a smaller circle with radius less than $1$ placed in the interior of the given triangle, which contradicts the fact that the given triangle has sides as arcs of circles of radius $1$. Therefore, each face of $P_1$, as well as $P_2, P_3,$ and $P_4$, is congruent to the given triangle. Since the sum of the angles of each pyramid is less than $2\pi$, and each pyramid has one face congruent to the given triangle, we can place four copies of the given triangle on the sphere such that they cover the entire surface. Therefore, four copies of such a triangle can cover the entire sphere.