During a rodeo, a clown runs 8.0 m north, turns 55 degrees 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?

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`Correct Answer: 13 m at 57 degrees North of East`

My Attempt:

x1 = 0 m

y1 = 8.0 m

x2 = [cos(55d)](3.5m) = 2 m

y2 = [sin(55d)](3.5m) = 2.9 m

total delta x = 2 m + 5 m = 7 m

total delta y = 10.9 m

c^2 = a^2 + b^2 = (7^2) + (10.9^2) = 49 + 118.81 = 167.81

c (hypotenuse) = 13 m (Ding! Thanks andrew)

tan theta (angle) = opposite / adjacent

theta (angle) = (opposite / adjacent) / tan

theta (angle) = ( 7 m / 10.9 m) / tan = 57 (Andrew, you rock.)

If anyone knows what I'm doing wrong (apparently a lot), feel free to post and tell me how stupid I am, haha.

EDIT: Fixed, thanks so much andrew.