What is the resultant displacement of 6m north, 8m east and 10m north west?

To find the resultant displacement, we can break down the given displacements into their respective components and then add them together.

The given displacements are:

6m north

8m east

10m northwest

Let's resolve the 10m northwest displacement into its north and west components. This can be done by using basic trigonometry. The northwest direction can be seen as a combination of north and west.

Using the Pythagorean theorem, we can find the magnitude of the northwest displacement:

Magnitude of northwest displacement = sqrt((10m)^2 + (10m)^2)

= sqrt(100m^2 + 100m^2)

= sqrt(200m^2)

= 10√2 m

To find the north and west components of the northwest displacement, we can use the sine and cosine ratios:

North component of northwest displacement = 10m * sin(45°)

= 10m * (1/√2)

= 10√2/2 m

= 5√2 m

West component of northwest displacement = 10m * cos(45°)

= 10m * (1/√2)

= 10√2/2 m

= 5√2 m

Now we have the following displacements:

6m north

8m east

5√2 m north (from the northwest displacement)

5√2 m west (from the northwest displacement)

To find the resultant displacement, we can add the north and south components separately and then add the east and west components separately:

North component = 6m + 5√2 m

= 6m + 5√2 m

East component = 8m

The final resultant displacement is:

Resultant displacement = North component + East component

= (6m + 5√2 m) north + 8m east