Solve for the remaining angles and side of the triangle described below. Round to the nearest thousandth: \[ B=100^{\circ}, c=2, a=1 \]

Step 1 ：We are given that in a triangle, angle B = 100 degrees, side c = 2 units, and side a = 1 unit. We are asked to find the remaining angles and side of the triangle.

Step 2 ：We can use the Law of Sines to solve for the remaining angles and side of the triangle. The Law of Sines states that the ratio of the length of a side of a triangle to the sine of its opposite angle is the same for all three sides of the triangle. This can be written as: \(\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}\)

Step 3 ：First, we can find angle A using the Law of Sines. We rearrange the formula to solve for A: \(A = \sin^{-1}(\frac{a \sin B}{c})\). Substituting the given values, we get: \(A = \sin^{-1}(\frac{1 \sin 100}{2})\). Solving this gives us \(A \approx 29.499^\circ\).

Step 4 ：Next, we can find angle C. Since the sum of the angles in a triangle is 180 degrees, we can subtract the known angles from 180 to find C: \(C = 180 - A - B\). Substituting the known values, we get: \(C = 180 - 29.499 - 100\). Solving this gives us \(C \approx 50.501^\circ\).

Step 5 ：Finally, we can find side b using the Law of Sines. We rearrange the formula to solve for b: \(b = \frac{c \sin B}{\sin A}\). Substituting the known values, we get: \(b = \frac{2 \sin 100}{\sin 29.499}\). Solving this gives us \(b \approx 1.567\).

Step 6 ：Final Answer: The remaining angles and side of the triangle are \(\boxed{A \approx 29.499^\circ, C \approx 50.501^\circ, b \approx 1.567}\).