smc

This solution refers to the Diagram section. Let $D$ be the orthogonal projection of $B$ onto the equator. Note that $\angle BDA=\angle BDC=90^{\circ }$ and $\angle BCD=45^{\circ }.$ Recall that $115^{\circ }\text{ W}$ longitude is the same as $245^{\circ }\text{ E}$ longitude, so $\angle ACD=135^{\circ }.$ We obtain the following diagram: Without loss of generality, let $AC=BC=1.$ For tetrahedron $ABCD:$ 1. Since $△BCD$ is an isosceles right triangle, we have $BD=CD=\frac{\sqrt{2}}{2}.$ 2. In $△ACD,$ we apply the Law of Cosines to get $AD=\sqrt{A{C}^2+C{D}^2-2\cdot AC\cdot CD\cdot \cos \angle ACD}=\frac{\sqrt{10}}{2}.$ 3. In right $△ABD,$ we apply the Pythagorean Theorem to get $AB=\sqrt{A{D}^2+B{D}^2}=\sqrt{3}.$ 4. In $△ABC,$ we apply the Law of Cosines to get $\cos \angle ACB=\frac{A{C}^2+B{C}^2-A{B}^2}{2\cdot AC\cdot BC}=-\frac{1}{2},$ so $\angle ACB=\boxed{120}$ degrees.