James stand at the centre of a regular field. he first take 50 steps North then 25 step West and finally 50 steps on the bearing of 315°. i. sketch James movement
ii. how far west is James final point from the centre?
iii. how far north is James final point from the centre?
iv. describe how you would guide a jhs student to find the bearing and distance of James final point from the centre.

Step-by-step explanation:

i. For navigation purposes, bearing is measured clockwise from north. In (x, y) coordinates, a distance D at a bearing B will have coordinates ...

  (x, y) = (Dsin(B), Dcos(B))

Then 50 steps north (bearing 0°) will put James at coordinates ...

  (x, y) = (50sin(0), 50cos(0)) = (0, 50)

The movement 25 steps west (bearing 270°) will add a displacement of ...

  (x, y) = (25sin(270°), 25cos(270°)) = (-25, 0)

Finally, the movement of 50 steps on bearing 315° will add a displacement of ...

  (x, y) = (50sin(315°), 50cos(315°)) = (-25√2, 25√2)

These movements are shown by the arrows to N, W, and F in the attached diagram.

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ii. James's final displacement is the sum of the individual displacements:

  (0, 50) +(-25, 0) +(-25√2, 25√2) = (-25(1+√2), 25(2+√2))

James is 25(1+√2) ≈ 60.4 steps west of center.

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iii. James is 25(2+√2) ≈ 85.4 steps north of center.

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iv. The distance can be found using the Pythagorean theorem (or distance formula). The distance from the origin to the final position (OF in the diagram) will be the root of the sum of the squares of the north and west displacements:

  distance = √(85.355² +60.355²)

  distance ≈ 104.5 steps

The bearing can be found using the arctangent function. The diagram shows you the reference angle (relative to the +y direction) has an opposite side equal to the west displacement, and an adjacent side equal to the north displacement. Then the bearing angle (β) will be ...

  tan(β) = opposite/adjacent = -60.355/85.355

  β ≈ arctan(-0.707106) ≈ -35.3°

The positive bearing angle is 360° added to this, or

  bearing = 324.7°