1. **Problem Statement:** Construct triangle ABC with given conditions: $AB=3$ cm, $\angle ABC=40^\circ$, and $\angle BAC=70^\circ$. 2. **Step 1: Understand the problem.** We are given one side length and two angles adjacent to that side. We need to construct the triangle using these. 3. **Step 2: Use the Triangle Angle Sum Rule.** The sum of interior angles in a triangle is always $180^\circ$. $$\angle ABC + \angle BAC + \angle BCA = 180^\circ$$ 4. **Step 3: Calculate the missing angle $\angle BCA$.** $$\angle BCA = 180^\circ - 40^\circ - 70^\circ = 70^\circ$$ 5. **Step 4: Use the Law of Sines to find other sides.** The Law of Sines states: $$\frac{AB}{\sin \angle BCA} = \frac{BC}{\sin \angle BAC} = \frac{AC}{\sin \angle ABC}$$ 6. **Step 5: Calculate side $BC$.** $$\frac{3}{\sin 70^\circ} = \frac{BC}{\sin 70^\circ} \Rightarrow BC = 3 \times \frac{\sin 70^\circ}{\sin 70^\circ} = 3$$ 7. **Step 6: Calculate side $AC$.** $$\frac{3}{\sin 70^\circ} = \frac{AC}{\sin 40^\circ} \Rightarrow AC = 3 \times \frac{\sin 40^\circ}{\sin 70^\circ}$$ Calculate numeric value: $$AC \approx 3 \times \frac{0.6428}{0.9397} \approx 3 \times 0.684 = 2.05 \text{ cm}$$ 8. **Step 7: Construction summary:** - Draw side $AB$ of length 3 cm. - At point $B$, construct an angle of $40^\circ$. - At point $A$, construct an angle of $70^\circ$. - The intersection of these two rays determines point $C$. --- **For questions 4, 5, and 6:** - Use the Triangle Angle Sum Rule to verify or find missing angles. - Use given side lengths and angles to apply the Law of Sines or Law of Cosines if needed. - Draw the triangle by starting with a known side or angle and constructing the other parts accordingly. This approach applies similarly to all these construction problems.