What Is the Person's Resultant Displacement After the Walk?

[chanv1]

Homework Statement

A person going for a walk follows the path shown in the figure, where y1 = 290 m and θ = 61.0°. The total trip consists of four straight-line paths. At the end of the walk, what is the person's resultant displacement measured from the starting point?

____ m? ____ degrees? (from the + x axis)

Homework Equations

http://img257.imageshack.us/img257/3813/physicsho5.th.jpg

The Attempt at a Solution

I've calculated and found the correct angle, which is 238.08 degrees but I don't know why I keep getting the wrong magnitude. I got 216.45 m and it's incorrect.

here is my work:

vector A = 100N @ 0 degrees ; A_x = 100 A_y = 0
vector B = 290N @ 270 degrees; B_x = 0 B_y = -290
vector C = 150N @ 210 degrees; C_x = -129.90 C_y = -75
vector D = 200N @ 115 degrees; D_x = -84.52 D_y = 181.26
____________________________________________________
F_x = -114.42; F_y = -183.74

tan^-1 (-183.74/-114.42) = 58.1 degrees + 180 degrees = 238.08 degrees

so for the magnitude I did

-114.42^2 + -183.74^2 = c^2
sq root of 46852.3 ... or 216.454 = c

but that is wrong, can anyone tell me what c would be? thanks.

 

[chanv1]

sorry, I've attached the picture.

The first vector is vector B in my solutions.

 

[Integral]

Ok, I"ve got the pic. The only problem I see is the value of the angle of the 4th vector. You have 115deg, I get 119deg.

 

[chanv1]

thank youuuu! how do I mark this as solved now?