1. The problem involves three triangles with given angles and side lengths. We need to analyze the first triangle with angles 90°, 40°, 50° and sides 13 cm, 5 cm, 12 cm. 2. Recall the Triangle Angle Sum Theorem: the sum of interior angles in a triangle is always 180°. 3. Check the first triangle's angles: $90^\circ + 40^\circ + 50^\circ = 180^\circ$, so the angles are valid. 4. Next, verify if the given sides correspond to these angles using the Law of Sines: $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$ where $a,b,c$ are sides opposite angles $A,B,C$ respectively. 5. Assign sides: let $a=13$ cm opposite $90^\circ$, $b=5$ cm opposite $40^\circ$, $c=12$ cm opposite $50^\circ$. 6. Calculate ratios: $$\frac{13}{\sin 90^\circ} = \frac{13}{1} = 13$$ $$\frac{5}{\sin 40^\circ} = \frac{5}{0.6428} \approx 7.78$$ $$\frac{12}{\sin 50^\circ} = \frac{12}{0.7660} \approx 15.67$$ 7. Since these ratios are not equal, the side lengths do not correspond to the given angles in a triangle. Final answer: The given sides 13 cm, 5 cm, and 12 cm do not form a triangle with angles 90°, 40°, and 50° according to the Law of Sines.