a man walks 8km North and then 5km in a. direction 60degrees. find its distance from his starting point

Use the component method to break down each vector into x,y components.

8km North = <0,8>
5km at 60° = <5cos60°,5sin60°>
Assuming angle is in Cartesian plane (counterclockwise from east) and not bearing (clockwise from north).

Add the two vectors and calculate the magnitude of the resultant.

56

5km

d = 8km[90o] + 5km[60o],

d = X + Yi = (8*Cos90+5*Cos60) + (8*sin90+5*sin60)I = 2.5 + 12.3i,
d = sqrt(X^2+Y^2) = 12.6 km.
Tan A = Y/x.

To find the distance from the starting point, we can use the concept of vector addition.

Let's break down the man's movement into two components: the northward movement and the movement at 60 degrees.

1. Northward movement: The man walks 8 km in the north direction. This can be represented as a vector pointing upwards by 8 km.

2. Movement at 60 degrees: The man walks 5 km in a direction 60 degrees from the north. To find the vector representation of this movement, we need to split it into its horizontal and vertical components. The vertical component can be found using sin(60°) = opposite/hypotenuse, where the opposite side is 5 km. Similarly, the horizontal component can be found using cos(60°) = adjacent/hypotenuse, where the adjacent side is 5 km.

Vertical component = 5 km * sin(60°) ≈ 4.33 km
Horizontal component = 5 km * cos(60°) = 2.5 km

Now, let's add the two vectors together to find the resultant displacement:

Vertical component: 8 km + 4.33 km = 12.33 km
Horizontal component: 0 km + 2.5 km = 2.5 km

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

Resultant magnitude = √(Vertical component^2 + Horizontal component^2)
= √(12.33 km^2 + 2.5 km^2)
≈ √(153.0089 km^2)
≈ 12.37 km

Therefore, the man's distance from his starting point is approximately 12.37 km.