Two sides and an angle are given below. Determine whether the given information results in one triangle, two triangles, or no triangle at all. Solve any resulting triangle(s).
$$b=10,c=7,B={120}^{\circ }$$

Step 1 ：Given the sides and angle of a triangle as $b=10$, $c=7$, and $B={120}^{\circ }$.

Step 2 ：Use the Law of Sines to determine whether the given information results in one triangle, two triangles, or no triangle at all. The Law of Sines states that the ratio of the length of a side of a triangle to the sine of the angle opposite that side is the same for all three sides of the triangle.

Step 3 ：Calculate the value of $\sin (C)$ using the formula: $\sin (C)=\sin (B)\cdot \frac{c}{b}$.

Step 4 ：Since $\sin (C)<1$, two triangles are possible.

Step 5 ：Calculate the measure of angle C for both possible triangles. The first possible measure of angle C is $\arcsin (\sin (C))$, and the second possible measure of angle C is $\pi -\arcsin (\sin (C))$.

Step 6 ：Find the measure of angle A for both triangles using the formula $A=\pi -B-C$.

Step 7 ：Find the length of side a for both triangles using the Law of Sines. The formula for side a is $a=b\cdot \frac{\sin (A)}{\sin (B)}$.

Step 8 ：Final Answer: The given information results in two possible triangles. The first triangle has sides $a=4.45$, $b=10$, and $c=7$, and angles $A={22.68}^{\circ }$, $B={120}^{\circ }$, and $C={37.32}^{\circ }$. The second triangle has sides $a=-11.45$, $b=10$, and $c=7$, and angles $A=-{82.68}^{\circ }$, $B={120}^{\circ }$, and $C={142.68}^{\circ }$. However, the negative values for side a and angle A in the second triangle indicate that this triangle is not possible in Euclidean geometry. Therefore, only the first triangle is a valid solution. $\boxed{{\displaystyle a=4.45,A={22.68}^{\circ },B={120}^{\circ },C={37.32}^{\circ }}}$