Trigonometry, bearings and elevation

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Explanation

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Worked solution

1. The hiker walks 27 m on a bearing of 045 degrees from point A to point B. This means the direction is 45 degrees from north, towards the east, forming a 45-degree angle with the north-south axis.
2. From point B, the hiker observes point A at a bearing of 225 degrees. A bearing of 225 degrees is 180 degrees opposite to 045 degrees, meaning point A lies directly southwest of point B.
3. The path from A to B and the return observation form a triangle ABC, where C is not needed. The angle at point B between the path AB and the line BA is 180 degrees, but the bearings indicate that the direction from B to A is 225 degrees, and from A to B is 045 degrees.
4. The angle between the two bearings (045° and 225°) at point B is 225° - 045° = 180°, but this is not the internal angle of the triangle. Instead, the internal angle at B is formed by the direction from B to A and from B to the original path. Since the hiker turned 180° to face back, the internal angle at B is 90 degrees.
5. The triangle formed by points A, B, and the direction of travel is a right-angled triangle at B, because the bearings 045° and 225° differ by 180°, and the path from A to B and back to A forms a straight line, but the hiker's turn implies a 90-degree internal angle due to the symmetry of the bearings.
6. The path from A to B is 27 m, and the angle at B is 90 degrees. The straight-line distance from A to B is the same as the path walked, which is 27 m. However, the question asks for the straight-line distance from A to B using trigonometry and bearings, which confirms that AB is the hypotenuse of a right-angled triangle with legs along the north-south and east-west axes.
7. Since the bearing is 045°, the path AB makes a 45-degree angle with both the north and east directions. Therefore, the eastward and northward components of AB are equal. Let x be the eastward (or northward) component. Then, by trigonometry: x = 27 * cos(45°) = 27 * (sqrt(2)/2).
8. The straight-line distance from A to B is the magnitude of the displacement vector. Since the hiker walked 27 m at 45°, the straight-line distance is simply 27 m, as it is the direct path. However, to confirm using Pythagoras: if the east and north components are both 27 * (sqrt(2)/2), then the distance is sqrt( (27*sqrt(2)/2)^2 + (27*sqrt(2)/2)^2 ).
9. Compute: (27*sqrt(2)/2)^2 = (729 * 2) / 4 = 1458 / 4 = 364.5. Then, sum of squares: 364.5 + 364.5 = 729. So, distance = sqrt(729) = 27 m.
10. The calculation confirms that the straight-line distance from A to B is exactly 27 m, as the path walked is direct and the bearings confirm a 45-degree angle, making the displacement equal to the path length.
11. The answer is 27 m, which is exact and matches the path length. No radical or fraction is needed because the distance walked is the straight-line distance.
12. Final verification: since the hiker walked 27 m on a bearing of 045°, and no detour or return path was taken, the straight-line distance from A to B is exactly 27 m.

Reference answer: 27 m

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