Answer
$(10.0~m)~\hat{i}+(1.62~m)~\hat{j}$
Work Step by Step
We can express each vector in unit-vector notation:
vector $P$:
$P_x = (10.0~m)~cos~25.0^{\circ} = 9.06~m$
$P_y = (10.0~m)~sin~25.0^{\circ} = 4.23~m$
$P = (9.06~m)~\hat{i}+(4.23~m)~\hat{j}$
vector $Q$:
$Q_x = -(12.0~m)~sin~10.0^{\circ} = -2.08~m$
$Q_y = (12.0~m)~cos~10.0^{\circ} = 11.8~m$
$Q = (-2.08~m)~\hat{i}+(11.8~m)~\hat{j}$
vector $R$:
$R_x = -(8.00~m)~sin~20.0^{\circ} = -2.74~m$
$R_y = -(8.00~m)~cos~20.0^{\circ} = -7.52~m$
$R = (-2.74~m)~\hat{i}+(-7.52~m)~\hat{j}$
vector $S$:
$S_x = (9.00~m)~sin~40.0^{\circ} = 5.79~m$
$S_y = -(9.00~m)~cos~40.0^{\circ} = -6.89~m$
$S = (5.79~m)~\hat{i}+(-6.89~m)~\hat{j}$
We can find the sum of the x components of the four vectors:
$P_x+Q_x+R_x+S_x = (9.06~m)+(-2.08~m)+(-2.74~m)+(5.79~m) = 10.0~m$
We can find the sum of the y components of the four vectors:
$P_y+Q_y+R_y+S_y = (4.23~m)+(11.8~m)+(-7.52~m)+(-6.89~m) = 1.62~m$
We can express the sum of the four vectors in unit-vector notation:
$(10.0~m)~\hat{i}+(1.62~m)~\hat{j}$
