1. **Problem statement:** Sally is located such that from checkpoint A she is N32.6°E, and from checkpoint B she is N15.0°W. The distance between checkpoints A and B is 45 km, with B directly east of A. We need to find Sally's distances from A and B, denoted as $x$ and $y$ respectively. 2. **Setup and known values:** - Distance $AB = 45$ km (horizontal line from west to east). - Angle at A to Sally: $32.6^\circ$ north of east. - Angle at B to Sally: $15.0^\circ$ north of west. 3. **Diagram description:** - Draw a horizontal line $AB$ with $A$ on the left and $B$ on the right, length 45 km. - From $A$, draw a line at $32.6^\circ$ above the horizontal to represent Sally's direction. - From $B$, draw a line at $15.0^\circ$ above the horizontal but slanting left (west) to represent Sally's direction. - The intersection of these two lines is Sally's position. 4. **Using trigonometry:** Let Sally's coordinates be $(X, Y)$ with $A$ at $(0,0)$ and $B$ at $(45,0)$. From $A$: $$\tan(32.6^\circ) = \frac{Y}{X} \implies Y = X \tan(32.6^\circ)$$ From $B$: $$\tan(15.0^\circ) = \frac{Y}{45 - X} \implies Y = (45 - X) \tan(15.0^\circ)$$ 5. **Equate the two expressions for $Y$:** $$X \tan(32.6^\circ) = (45 - X) \tan(15.0^\circ)$$ 6. **Solve for $X$:** $$X \tan(32.6^\circ) = 45 \tan(15.0^\circ) - X \tan(15.0^\circ)$$ $$X \tan(32.6^\circ) + X \tan(15.0^\circ) = 45 \tan(15.0^\circ)$$ $$X (\tan(32.6^\circ) + \tan(15.0^\circ)) = 45 \tan(15.0^\circ)$$ $$X = \frac{45 \tan(15.0^\circ)}{\tan(32.6^\circ) + \tan(15.0^\circ)}$$ 7. **Calculate values:** $$\tan(32.6^\circ) \approx 0.639$$ $$\tan(15.0^\circ) \approx 0.268$$ $$X = \frac{45 \times 0.268}{0.639 + 0.268} = \frac{12.06}{0.907} \approx 13.29 \text{ km}$$ 8. **Calculate $Y$:** $$Y = X \tan(32.6^\circ) = 13.29 \times 0.639 \approx 8.49 \text{ km}$$ 9. **Find distances $x$ and $y$ (Sally to A and B):** - Distance from A: $$x = \sqrt{X^2 + Y^2} = \sqrt{13.29^2 + 8.49^2} = \sqrt{176.7 + 72.1} = \sqrt{248.8} \approx 15.77 \text{ km}$$ - Distance from B: Coordinates of B are $(45,0)$, so $$y = \sqrt{(45 - X)^2 + Y^2} = \sqrt{(45 - 13.29)^2 + 8.49^2} = \sqrt{31.71^2 + 8.49^2} = \sqrt{1005.5 + 72.1} = \sqrt{1077.6} \approx 32.83 \text{ km}$$ **Final answer:** Sally is approximately 15.77 km from checkpoint A and 32.83 km from checkpoint B.