A man starts from a Point a and walks 4km on a bearing of 015. he then walks 6km on a bearing of 105 to c. What is the bearing of c from a.
To find the bearing of point C from point A, we need to use trigonometry and vector addition.
Let's break down the information given step by step:
1. The man starts at point A.
2. He walks 4 km on a bearing of 015 (in degrees).
3. He then walks 6 km on a bearing of 105.
To find the end position, we need to calculate the displacement vector for each leg of the journey and then sum them up.
Step 1: Finding the displacement vector for the first leg (4 km on a bearing of 015).
To calculate the x and y components of this displacement vector, we can use trigonometry.
The bearing of 015 can also be expressed as an angle relative to the positive x-axis, which is 360 - 15 = 345 degrees.
The x-component of the displacement is given by: 4 km * cos(345°)
The y-component of the displacement is given by: 4 km * sin(345°)
So, the displacement vector for the first leg is: (4 km * cos(345°), 4 km * sin(345°))
Step 2: Finding the displacement vector for the second leg (6 km on a bearing of 105).
Again, we can use trigonometry to calculate the x and y components of this displacement vector.
The x-component of the displacement is given by: 6 km * cos(105°)
The y-component of the displacement is given by: 6 km * sin(105°)
So, the displacement vector for the second leg is: (6 km * cos(105°), 6 km * sin(105°))
Step 3: Adding the two displacement vectors.
To get the total displacement vector, we need to add the x and y components of the two displacement vectors.
The x-component of the total displacement is: (4 km * cos(345°)) + (6 km * cos(105°))
The y-component of the total displacement is: (4 km * sin(345°)) + (6 km * sin(105°))
Now, we have the x and y components of the total displacement vector.
Step 4: Finding the bearing from point A to point C.
To find the bearing of point C from point A, we need to convert the total displacement vector back into a bearing angle.
The bearing angle can be calculated using the formula: bearing = atan2(y-component, x-component)
So, the bearing of point C from point A is: bearing = atan2((4 km * sin(345°)) + (6 km * sin(105°)), (4 km * cos(345°)) + (6 km * cos(105°)))
he walks on a heading, not a bearing.
A-to-B = (4cos75°,4sin75°) = (1.03,3.86)
B-toC = (6cos15°,-6sin15°) = (5.80,-1.55)
so, A-to-C = (6.83,2.31)
Now,
tanθ = 2.31/6.83
θ = 18.7°
so, the bearing of C from A is 90-θ = 71.3°