Which of the following can be drawn as triangles? Why or why not? a. \( \angle \mathrm{P}=70^{\circ}, \angle \mathrm{O}=40^{\circ} \) and \( \overline{0}=4 \mathrm{~cm} \)

Yes, a triangle **can** be drawn with the given measurements. Here's why: ### Given: - **Angle P** = \(70^{\circ}\) - **Angle O** = \(40^{\circ}\) - **Side \(\overline{O}\)** = 4 cm ### Reasoning: 1. **Sum of Angles:** - In any triangle, the sum of the interior angles must be \(180^{\circ}\). - Given: \[ \angle P + \angle O = 70^{\circ} + 40^{\circ} = 110^{\circ} \] - **Remaining Angle:** \[ \angle Q = 180^{\circ} - 110^{\circ} = 70^{\circ} \] - All three angles (\(70^{\circ}\), \(40^{\circ}\), and \(70^{\circ}\)) are positive and add up to \(180^{\circ}\), satisfying the triangle angle sum property. 2. **Side-Angle Relationship:** - With two angles known, the triangle is uniquely determined up to similarity (by the Angle-Angle (AA) postulate). - The given side (\(\overline{O} = 4 \text{ cm}\)) corresponds to one of the angles. Using the Law of Sines, the lengths of the other sides can be determined, ensuring consistency and feasibility. 3. **Law of Sines (for verification):** - The Law of Sines states: \[ \frac{\sin P}{\overline{p}} = \frac{\sin O}{\overline{o}} = \frac{\sin Q}{\overline{q}} \] - Plugging in the known values: \[ \frac{\sin 70^{\circ}}{\overline{p}} = \frac{\sin 40^{\circ}}{4 \text{ cm}} \] - This allows for the calculation of the other sides (\(\overline{p}\) and \(\overline{q}\)), confirming that the side lengths are proportional and the triangle is valid. ### Conclusion: Since the sum of the provided angles is less than \(180^{\circ}\) and the given side length is consistent with these angles, **a valid triangle can be constructed** with the specified measurements. **Answer:** Yes. The two given angles add up to less than 180°, allowing a third angle to exist, and the provided side length is consistent with these angles. Therefore, a triangle with angle P = 70°, angle O = 40°, and side Ō = 4 cm can be drawn.