What is the magnitude of the persons resultant displacement

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#34, 35

find the x and y components of each segment of the trip; call positive x to the right and positive y up:

x1 means x component of first segment, y2 means y component of second segement, etc...

x1 = + 108; y1 =0
x2 =0; y2 = -434
x3 = -287 cos 70 = - 98.2; y3 = - 287 sin 70 = - 269.7
x4 = - 302 cos 29 = - 264.1; y4 = + 302 sin 29 = + 146.4

add all the x's to find the total x component, similarly add all the y's:

total x component = -254.3; total y component = -557.3

total displacement = Sqrt[(-254.3)^2 +(-557.3)^2] = 612.6

resultant vector is in third quadrant, and makes an angle with respect to the negative x axis of

tan(theta) = 557/254 => theta = 65.4 deg...but this is the angle down from the negative x axis...the angle counterclockwise from the positive x axis is 180 deg + 65.4 deg = 245.4 deg

thank you for your time .. bottom answer!

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v= +108 i -434 j - 287 cos(70) i - 287 sin(70) j + 302 sin(29) j - 302 cos(29)i


|v |= sqrt ( (108- 287 cos(70) - 302 cos(29))^2 + (+ 302 sin(29) - 287 sin(70) -434)^2 )


theta = arctan ( ( 302 sin(29) - 287 sin(70) -434) / (108- 287 cos(70) - 302 cos(29)))