1. **Problem statement:** Find the measure of angle ABC in triangle ABC with points A(-2,4), B(3,7), and C(3,4) using trigonometry. 2. **Recall:** Angle ABC is the angle at point B formed by points A and C. 3. **Vectors:** To find angle ABC, consider vectors $ \overrightarrow{BA} $ and $ \overrightarrow{BC} $. \[ \overrightarrow{BA} = A - B = (-2 - 3, 4 - 7) = (-5, -3) \] \[ \overrightarrow{BC} = C - B = (3 - 3, 4 - 7) = (0, -3) \] 4. **Formula for angle between vectors:** $$ \cos \theta = \frac{\overrightarrow{BA} \cdot \overrightarrow{BC}}{|\overrightarrow{BA}| |\overrightarrow{BC}|} $$ where $ \cdot $ is the dot product and $ |\cdot| $ is the vector magnitude. 5. **Calculate dot product:** $$ \overrightarrow{BA} \cdot \overrightarrow{BC} = (-5)(0) + (-3)(-3) = 0 + 9 = 9 $$ 6. **Calculate magnitudes:** $$ |\overrightarrow{BA}| = \sqrt{(-5)^2 + (-3)^2} = \sqrt{25 + 9} = \sqrt{34} $$ $$ |\overrightarrow{BC}| = \sqrt{0^2 + (-3)^2} = \sqrt{9} = 3 $$ 7. **Calculate cosine of angle:** $$ \cos \theta = \frac{9}{\sqrt{34} \times 3} = \frac{9}{3\sqrt{34}} = \frac{3}{\sqrt{34}} $$ 8. **Calculate angle:** $$ \theta = \cos^{-1} \left( \frac{3}{\sqrt{34}} \right) $$ Using a calculator, $$ \theta \approx 57^\circ $$ **Final answer:** The measure of angle ABC is approximately **57 degrees**.