Answer
The magnitude of the displacement is $~~8.43~m$
Work Step by Step
We can find the components of the displacement for $d_1$:
$x_1 = (5.00~m)~cos~30.0^{\circ} = 4.33~m$
$y_1 = (5.00~m)~sin~30.0^{\circ} = 2.50~m$
Note that $d_2$ is at an angle of $20.0^{\circ}$ above the horizontal.
We can find the components of the displacement for $d_2$:
$x_2 = -(8.00~m)~cos~20.0^{\circ} = -7.52~m$
$y_2 = (8.00~m)~sin~20.0^{\circ} = 2.74~m$
Note that $d_3$ is at an angle of $10.0^{\circ}$ clockwise from the negative direction of the y axis.
We can find the components of the displacement for $d_3$:
$x_3 = -(12.0~m)~sin~10.0^{\circ} = -2.08~m$
$y_3 = -(12.0~m)~cos~10.0^{\circ} = -11.82~m$
We can find the net displacement:
$\Delta x = 4.33~m-7.52~m-2.08~m = -5.27~m$
$\Delta y = 2.50~m+2.74~m-11.82~m = -6.58~m$
We can find the magnitude of the displacement:
$d = \sqrt{(-5.27~m)^2+(-6.58~m)^2} = 8.43~m$
The magnitude of the displacement is $~~8.43~m$
