1. **Problem Statement:** Find the measures of the angles of the triangle with vertices A(-1,0), B(2,1), and C(1,-2). 2. **Formula and Rules:** To find the angles of a triangle given coordinates, we use the distance formula to find side lengths and then the Law of Cosines to find angles. - Distance formula between points $P_1(x_1,y_1)$ and $P_2(x_2,y_2)$: $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ - Law of Cosines for angle $\theta$ opposite side $a$: $$\cos \theta = \frac{b^2 + c^2 - a^2}{2bc}$$ 3. **Calculate side lengths:** - $AB = \sqrt{(2 - (-1))^2 + (1 - 0)^2} = \sqrt{3^2 + 1^2} = \sqrt{9 + 1} = \sqrt{10}$ - $BC = \sqrt{(1 - 2)^2 + (-2 - 1)^2} = \sqrt{(-1)^2 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10}$ - $AC = \sqrt{(1 - (-1))^2 + (-2 - 0)^2} = \sqrt{2^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8}$ 4. **Find angles using Law of Cosines:** - Angle at A (opposite BC): $$\cos A = \frac{AB^2 + AC^2 - BC^2}{2 \times AB \times AC} = \frac{10 + 8 - 10}{2 \times \sqrt{10} \times \sqrt{8}} = \frac{8}{2 \times \sqrt{80}} = \frac{8}{2 \times 8.944} = \frac{8}{17.888} \approx 0.447$$ $$A = \cos^{-1}(0.447) \approx 63.3^\circ$$ - Angle at B (opposite AC): $$\cos B = \frac{AB^2 + BC^2 - AC^2}{2 \times AB \times BC} = \frac{10 + 10 - 8}{2 \times \sqrt{10} \times \sqrt{10}} = \frac{12}{2 \times 10} = \frac{12}{20} = 0.6$$ $$B = \cos^{-1}(0.6) \approx 53.13^\circ$$ - Angle at C (opposite AB): $$C = 180^\circ - A - B = 180^\circ - 63.3^\circ - 53.13^\circ = 63.57^\circ$$ 5. **Final answer:** The angles of the triangle are approximately: - $A = 63.3^\circ$ - $B = 53.13^\circ$ - $C = 63.57^\circ$