Question 1199997: During a rodeo, a clown runs 8.0 m north, turns 55° north of east, and runs
3.5 m. Then, after waiting for the bull to come near, the clown turns due east
and runs 5.0 m to exit the arena. What is the clown’s total displacement?
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Hint:
This is one way to draw out the diagram
Segment BC may or may not be parallel to segment AD.
The rodeo clown follows the path from A, to B, to C, to D in that exact order.
A to B = go north 8.0 meters
B to C = travel 3.5 meters at an angle 55 degrees north of east
C to D = go east 5.0 meters
This path ABCD is marked in black.
The blue segments form right triangle BEC, where the 90 degree angle is at point E.
Let
x = length of segment EC
y = length of segment EB
You can use the trig ratios sine or cosine determine the approximate values of x and y.
sin(angle) = opposite/hypotenuse
cos(angle) = adjacent/hypotenuse
Example:
sin(angle) = opposite/hypotenuse
sin(angle EBC) = x/3.5
sin(angle EBC) = EC/BC
sin(35) = x/3.5
I'll let the student solve for x.
Make sure your calculator is in degree mode.
Determining the value of y involves a similar calculation.
After determining x and y, note how:
segment AE = AB+EB = 8.0 + y
segment ED = EC+CD = x + 5.0
Meaning we can determine the legs of right triangle AED.
The hypotenuse of triangle AED, shown in red, is the displacement vector.
It's the straight line path that the clown would take if the clown went directly from start to finish. It's the shortest such path.
This means the clown avoids any detours that waste time.
To find the length of the red vector, use the pythagorean theorem 
a & b are the leg lengths in either order (sides AE and ED calculated earlier)
c = hypotenuse = longest side = segment AD
I'll let the student do these calculations.
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