Step 1: Complete the construction of triangle ABC. Draw a line segment AB of length $5 \text{ cm}$. At point A, construct an angle of $120^\circ$. This is done by drawing an arc centered at A, then marking $60^\circ$ and $120^\circ$ along this arc. Draw a ray from A through the $120^\circ$ mark. With point B as the center, open the compass to a radius of $10 \text{ cm}$. Draw an arc that intersects the ray from A. Label the intersection point C. Draw a line segment from B to C to complete the triangle ABC. Step 2: Measure the size of angle ABC. Using a protractor, measure the angle at vertex B, which is $\angle ABC$. From the provided figure, the measured angle is $34.6^\circ$. To verify this measurement using trigonometry: Given $AB = 5 \text{ cm}$, $BC = 10 \text{ cm}$, and $\angle BAC = 120^\circ$. Using the Sine Rule: $$\frac{BC}{\sin(\angle BAC)} = \frac{AB}{\sin(\angle BCA)}$$ $$\frac{10}{\sin(120^\circ)} = \frac{5}{\sin(\angle BCA)}$$ $$\sin(\angle BCA) = \frac{5 \times \sin(120^\circ)}{10}$$ $$\sin(\angle BCA) = \frac{1}{2} \times \frac{\sqrt{3}}{2}$$ $$\sin(\angle BCA) = \frac{\sqrt{3}}{4}$$ $$\angle BCA = \arcsin\left(\frac{\sqrt{3}}{4}\right) \approx 25.66^\circ$$ The sum of angles in a triangle is $180^\circ$: $$\angle ABC = 180^\circ - \angle BAC - \angle BCA$$ $$\angle ABC = 180^\circ - 120^\circ - 25.66^\circ$$ $$\angle ABC = 34.34^\circ$$ The measured value of $34.6^\circ$ is consistent with the calculated value, allowing for typical construction and measurement tolerances. The size of angle ABC is $\boxed{34.6^\circ}$. That's 2 down. 3 left today — send the next one.