1. **Problem statement:** Anna and Julia start at point P. Point Q is directly north of P. Anna walks first on a bearing of 330°, then changes to 040° to reach Q. Julia walks 3 km on a bearing of 020°, then changes to 300° to reach Q. We need to find who took the shorter route. 2. **Understanding bearings:** Bearings are measured clockwise from north. 0° is north, 90° east, 180° south, 270° west. 3. **Set coordinate system:** Place P at origin $(0,0)$, Q lies on positive y-axis at $(0,y)$ for some $y$. 4. **Anna's path:** Let Anna's first segment length be $a$, second segment length be $b$. - First segment bearing 330° means direction vector components: $x_1 = a \sin 330^\circ = a \times (-\frac{1}{2}) = -\frac{a}{2}$, $y_1 = a \cos 330^\circ = a \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}a}{2}$. - Second segment bearing 040°: $x_2 = b \sin 40^\circ$, $y_2 = b \cos 40^\circ$. - Total displacement must be $(0,y)$ since Anna ends at Q directly north of P: $$x_1 + x_2 = 0 \implies -\frac{a}{2} + b \sin 40^\circ = 0$$ $$y_1 + y_2 = y \implies \frac{\sqrt{3}a}{2} + b \cos 40^\circ = y$$ 5. **Julia's path:** First segment length is 3 km on bearing 020°. - First segment components: $x_3 = 3 \sin 20^\circ$, $y_3 = 3 \cos 20^\circ$. - Second segment length $c$, bearing 300°: $x_4 = c \sin 300^\circ = c \times (-\frac{\sqrt{3}}{2}) = -\frac{\sqrt{3}c}{2}$ $y_4 = c \cos 300^\circ = c \times \frac{1}{2} = \frac{c}{2}$ - Total displacement must be $(0,y)$: $$x_3 + x_4 = 0 \implies 3 \sin 20^\circ - \frac{\sqrt{3}c}{2} = 0$$ $$y_3 + y_4 = y \implies 3 \cos 20^\circ + \frac{c}{2} = y$$ 6. **Solve for Anna:** From $x$-component: $$b = \frac{a/2}{\sin 40^\circ} = \frac{a}{2 \sin 40^\circ}$$ Substitute into $y$-component: $$y = \frac{\sqrt{3}a}{2} + \left(\frac{a}{2 \sin 40^\circ}\right) \cos 40^\circ = a \left(\frac{\sqrt{3}}{2} + \frac{\cos 40^\circ}{2 \sin 40^\circ}\right)$$ Calculate numeric values: $\sin 40^\circ \approx 0.6428$, $\cos 40^\circ \approx 0.7660$, $\sqrt{3}/2 \approx 0.8660$ $$y = a \left(0.8660 + \frac{0.7660}{2 \times 0.6428}\right) = a \left(0.8660 + \frac{0.7660}{1.2856}\right) = a (0.8660 + 0.596) = a (1.462)$$ So, $$a = \frac{y}{1.462}$$ Then, $$b = \frac{a}{2 \times 0.6428} = \frac{y/1.462}{1.2856} = \frac{y}{1.462 \times 1.2856} = \frac{y}{1.88}$$ 7. **Anna's total distance:** $$D_A = a + b = \frac{y}{1.462} + \frac{y}{1.88} = y \left(\frac{1}{1.462} + \frac{1}{1.88}\right) = y (0.684 + 0.532) = 1.216 y$$ 8. **Solve for Julia:** From $x$-component: $$3 \sin 20^\circ = \frac{\sqrt{3} c}{2} \implies c = \frac{2 \times 3 \sin 20^\circ}{\sqrt{3}} = \frac{6 \times 0.3420}{1.732} = \frac{2.052}{1.732} = 1.184$$ From $y$-component: $$y = 3 \cos 20^\circ + \frac{c}{2} = 3 \times 0.9397 + \frac{1.184}{2} = 2.819 + 0.592 = 3.411$$ 9. **Julia's total distance:** $$D_J = 3 + c = 3 + 1.184 = 4.184$$ 10. **Compare distances:** Anna's distance is $1.216 y$, Julia's is $4.184$. Since $y=3.411$ from Julia's calculation, $$D_A = 1.216 \times 3.411 = 4.147$$ Anna's route is approximately 4.147 km, Julia's is 4.184 km. **Answer:** Anna took the shorter route.