Bearing; a girl walks 50m on a bearing 025degrees and then 200m due east. how far is she from her starting point
North dist = 50 cos 25 = 45.3
East dist = 200 + 50 sin 25 = 200 + 21.1 = 221.1
dist = sqrt (45.3^2 + 221.1^2)
By the way you walk, sail, or fly on a heading. You take a bearing on something like a lighthouse giving the direction from you to it.
(Teach navigation, fussy about this)
The difference between heading and bearing becomes important in real life.
If you need to go around a bell buoy whose bearing from you is North, you only head North if there is no cross current or wind or other factor causing your heading not to be the same as the direction the boat or plane or whatever actually moves. I am convinced that people who write math texts do not drive ships or planes.
To find the distance from the girl's starting point, we can use the concept of vector addition.
First, let's break down the girl's movement into two components:
1. The distance she walks on a bearing of 025 degrees, which we'll call vector A.
2. The distance she walks due east, which we'll call vector B.
To calculate vector A, we can use trigonometry. The bearing of 025 degrees means that the angle between vector A and the north direction is 025 degrees. Since she walks 50m in this direction, we can use the sine and cosine functions to find the components of vector A:
- The eastward component (Ax) = 50 * cos(25°)
- The northward component (Ay) = 50 * sin(25°)
Next, vector B is simply a distance due east, so its eastward component (Bx) is 200m, and the northward component (By) is 0m.
To find the total displacement from the starting point, we add the corresponding components of vector A and vector B:
- The eastward component of the total displacement (Dx) = Ax + Bx
- The northward component of the total displacement (Dy) = Ay + By
Now, we can calculate the magnitude (or length) of the total displacement using the Pythagorean theorem:
- Total displacement (D) = sqrt(Dx^2 + Dy^2)
Let's calculate the total displacement to find the distance from the starting point:
Ax = 50 * cos(25°) ≈ 45.08m
Ay = 50 * sin(25°) ≈ 21.43m
Bx = 200m
By = 0m
Dx = Ax + Bx ≈ 45.08m + 200m ≈ 245.08m
Dy = Ay + By ≈ 21.43m + 0m ≈ 21.43m
D = sqrt(Dx^2 + Dy^2) ≈ sqrt(245.08^2 + 21.43^2) ≈ 246.65m
Therefore, she is approximately 246.65 meters away from her starting point.