Answer the following prompts with numerical answers only. If necessary, round answers to the nearest tenth of a unit. The diamet
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Answer:
Let's define East as the positive x-axis and North as the positive y-axis.
If Michelle's initial position is (0, 0)m
We know that Nathasha is 50m due East of Michelle.
Then the position of Natasha is (50, 0)m
Now we know that Natasha walks 20m due North, then the new position of her's is:
(50, 0 + 20)m = (50, 20)m
While Michelle walks 10m due South (South would be the negative y-axis, then we subtract 10 meters)
Michelle's new position will be:
(0, 0 - 10)m = (0, -10)m
Now we want to know the distance and bearing of Michelle from Natasha.
First, remember that the distance between two points (a, b) and (c, d) is given by:
Distance = √( (a - c)^2 + (b - d)^2)
Then the distance between Michelle and Natasha is:
Distance = √( (50m - 0m)^2 + (20m - (-10m))^2)
Distance = √( (50m)^2 + (30m)^2) = 58.31m
Now to find the bearing you can see the image below:
Point B is Michelle's position and point A is Natasha's position.
To find the bearing, we can make a triangle rectangle as the one shown in the image:
Also remember the relation:
Tan(a) = (opposite cathetus)/(adjacent cathetus).
Where the opposite cathetus is the difference between the x-values of each position, this is:
opposite cathetus = 50m - 0m = 50m
And the adjacent cathetus is the difference between the y-values, this is:
adjacent cathetus = 20m - (-10m) = 30m
Then:
Tan(a) = 50m/30m
Tan(a) = 5/3
Now if we apply the inverse Tan function to both sides, Atan(x) we get:
Atan(Tan(a)) = Atan(5/3)
a = Atan(5/3) = 59°
So the bearing is of 59°.
