A walks 2Km south to B, she the walks 3km west to C. Find the distance and bearing from C and A
Pythagorean Theorem
a^2 + b^2 = c^2
2^2 + 3^2 = c^2
4 + 9 = c^2
13 = c^2
3.61 = c
This questions is like a word problem for a triangle. Draw point A and head 2 unit down draw point B there, then draw a point 3 units to the right. draw point C there it should be a right triangle with C-A as the hypotenuse. So distance C-A equals the square root of 2^2 + 3^2 using the pythagorean theorem hope this helps.
5KM 236 DEGREES
U are stupid
To find the distance and bearing from point C to point A, we can use the concept of vector addition and trigonometry.
First, let's draw a diagram to visualize the situation:
B
|
|
A---C
Given that A walks 2 km south to reach B and then 3 km west to reach C, we can represent these displacements as vectors.
Vector AB: 2 km south
Vector BC: 3 km west
To find the displacement between A and C, we can add these two vectors together:
AC = AB + BC
To add these vectors, we can break them down into their x (east-west) and y (north-south) components.
For vector AB:
- x component = 0 km (since it's purely southward)
- y component = -2 km (negative because it's southward)
For vector BC:
- x component = -3 km (negative because it's westward)
- y component = 0 km (since it's purely westward)
Adding these components together, we get:
ACx = 0 km + (-3 km) = -3 km
ACy = -2 km + 0 km = -2 km
Now, let's calculate the distance between A and C using the Pythagorean theorem:
Distance AC = √(ACx^2 + ACy^2)
= √((-3 km)^2 + (-2 km)^2)
= √(9 km^2 + 4 km^2)
= √(13 km^2)
≈ 3.61 km
Therefore, the distance from C to A is approximately 3.61 km.
To calculate the bearing from C to A, we can use trigonometry. The bearing is usually measured clockwise from the north direction.
Bearing = arctan(ACy / ACx)
Bearing = arctan((-2 km) / (-3 km))
= arctan(2/3)
≈ 33.69 degrees
Therefore, the bearing from C to A is approximately 33.69 degrees clockwise from the north direction.