1. **Problem Statement:** Measure the angles and calculate the side lengths of triangles ABC and PQR where $A=(0,0), B=(4,-3), C=(6,3), P=(-2,7), Q=(9,5), R=(7,19)$. 2. **Formulas and Rules:** - Distance between points $X=(x_1,y_1)$ and $Y=(x_2,y_2)$ is given by $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ - To find angles, use the Law of Cosines: $$\cos \theta = \frac{a^2 + b^2 - c^2}{2ab}$$ where $a,b,c$ are side lengths opposite to angles. 3. **Calculate side lengths of triangle ABC:** - $AB = \sqrt{(4-0)^2 + (-3-0)^2} = \sqrt{16 + 9} = \sqrt{25} = 5$ - $BC = \sqrt{(6-4)^2 + (3+3)^2} = \sqrt{2^2 + 6^2} = \sqrt{4 + 36} = \sqrt{40} = 2\sqrt{10}$ - $AC = \sqrt{(6-0)^2 + (3-0)^2} = \sqrt{36 + 9} = \sqrt{45} = 3\sqrt{5}$ 4. **Calculate angles of triangle ABC:** - Angle at $A$ opposite side $BC$: $$\cos \angle A = \frac{AB^2 + AC^2 - BC^2}{2 \times AB \times AC} = \frac{5^2 + (3\sqrt{5})^2 - (2\sqrt{10})^2}{2 \times 5 \times 3\sqrt{5}}$$ Calculate numerator: $25 + 45 - 40 = 30$ Calculate denominator: $2 \times 5 \times 3\sqrt{5} = 30\sqrt{5}$ So, $$\cos \angle A = \frac{30}{30\sqrt{5}} = \frac{1}{\sqrt{5}}$$ Therefore, $$\angle A = \cos^{-1}\left(\frac{1}{\sqrt{5}}\right) \approx 63.43^\circ$$ - Angle at $B$ opposite side $AC$: $$\cos \angle B = \frac{AB^2 + BC^2 - AC^2}{2 \times AB \times BC} = \frac{25 + 40 - 45}{2 \times 5 \times 2\sqrt{10}} = \frac{20}{20\sqrt{10}} = \frac{1}{\sqrt{10}}$$ $$\angle B = \cos^{-1}\left(\frac{1}{\sqrt{10}}\right) \approx 71.57^\circ$$ - Angle at $C$: $$\angle C = 180^\circ - \angle A - \angle B = 180^\circ - 63.43^\circ - 71.57^\circ = 45^\circ$$ 5. **Calculate side lengths of triangle PQR:** - $PQ = \sqrt{(9+2)^2 + (5-7)^2} = \sqrt{11^2 + (-2)^2} = \sqrt{121 + 4} = \sqrt{125} = 5\sqrt{5}$ - $QR = \sqrt{(7-9)^2 + (19-5)^2} = \sqrt{(-2)^2 + 14^2} = \sqrt{4 + 196} = \sqrt{200} = 10\sqrt{2}$ - $PR = \sqrt{(7+2)^2 + (19-7)^2} = \sqrt{9^2 + 12^2} = \sqrt{81 + 144} = \sqrt{225} = 15$ 6. **Calculate angles of triangle PQR:** - Angle at $P$ opposite side $QR$: $$\cos \angle P = \frac{PQ^2 + PR^2 - QR^2}{2 \times PQ \times PR} = \frac{125 + 225 - 200}{2 \times 5\sqrt{5} \times 15} = \frac{150}{150\sqrt{5}} = \frac{1}{\sqrt{5}}$$ $$\angle P = \cos^{-1}\left(\frac{1}{\sqrt{5}}\right) \approx 63.43^\circ$$ - Angle at $Q$ opposite side $PR$: $$\cos \angle Q = \frac{PQ^2 + QR^2 - PR^2}{2 \times PQ \times QR} = \frac{125 + 200 - 225}{2 \times 5\sqrt{5} \times 10\sqrt{2}} = \frac{100}{100\sqrt{10}} = \frac{1}{\sqrt{10}}$$ $$\angle Q = \cos^{-1}\left(\frac{1}{\sqrt{10}}\right) \approx 71.57^\circ$$ - Angle at $R$: $$\angle R = 180^\circ - \angle P - \angle Q = 180^\circ - 63.43^\circ - 71.57^\circ = 45^\circ$$ 7. **Conclusion:** Triangles ABC and PQR have the same side length ratios and the same angles: approximately $63.43^\circ$, $71.57^\circ$, and $45^\circ$. Therefore, triangles ABC and PQR are similar by Angle-Angle similarity criterion.