a man walks 8km north and 5km in a direction east of north.find the distance from the starting point
8kmN moves (0,8)
5km NE moves (3.5,3.5)
add them up to get (3.5,11.5)
distance is thus √(3.5^2 + 11.5^2) = 12.0
1. A man walks 10km east and 20km south the displacement of the man from initial position
2. In figure below R= 30, find x and Y respectively.
What is the length of the hypotenuse? If necessary, round to the nearest tenth.
To find the distance from the starting point, we can use the Pythagorean theorem.
First, let's visualize the man's movements. He walks 8km north and then 5km east of north. We can draw a right triangle to represent this situation. The 8km north movement forms the vertical side (opposite side), and the 5km east of north movement forms the horizontal side (adjacent side).
Now, we can use the Pythagorean theorem, which states that in a right triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.
In this case, the hypotenuse represents the distance from the starting point, and we need to find its length. Let's call this length "d."
Using the Pythagorean theorem:
d^2 = (8km)^2 + (5km)^2
Simplifying:
d^2 = 64km^2 + 25km^2
d^2 = 89km^2
To find the value of "d," we take the square root of both sides:
d = √89km
So, the distance from the starting point is approximately 9.43 km (rounded to two decimal places).
This is a pythagorean theorem problem:
a^2 + b^2 = c^2
You have two sides and you are going to solve for c.