Answer
$(2.1m)\hat x+(0.74m)\hat y$
Work Step by Step
First of all, we find the x and y components of the given vectors
$A_x=Acos\theta$
$A_x=(1.5m)cos40^{\circ}=1.1m$
$B_x=Bcos\theta$
$B_x=(2.0m)cos(-19)^{\circ}=1.9m$
$C_x=Ccos\theta$
$C_x=(1.0m)cos(180-25)^{\circ}=-0.91m$
$A_x+B_x+C_x=1.1+1.9-0.91=2.1m$
Now, we find the y components of the given vectors:
$A_y=Asin\theta$
$A_y=(1.5m)sin40^{\circ}=0.96m$
$B_y=Bsin\theta$
$B_y=(2.0m)sin(-19)^{\circ}=-0.65m$
$C_y=Csin\theta$
$C_y=(1.0m)sin(180-25)^{\circ}=0.42m$
Now we can find $\vec A+\vec B+\vec C$
$\vec A+\vec B+\vec C=(2.1m)\hat x+(0.74)\hat y$
