Question 1206729
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Let's say the 160 newton force is pulling directly eastward, and the other vector points in the northeast direction.


For the 160 newton vector
x = r*cos(theta) = 160*cos(0) = 160
y = r*sin(theta) = 160*sin(0) = 0
The < x,y > form of this vector is < 160,0 >


For the 280 newton vector
x = r*cos(theta) = 280*cos(45) = 280*sqrt(2)/2 = 140*sqrt(2)
y = r*sin(theta) = 280*sin(45) = 280*sqrt(2)/2 = 140*sqrt(2)
The < x,y > form of this vector is < 140*sqrt(2), 140*sqrt(2) >


The two vectors are
< 160,0 >
< 140*sqrt(2), 140*sqrt(2) >
Add straight down to get the resultant vector in component form.
< 160+140*sqrt(2), 140*sqrt(2) >


That approximates to roughly,
< 357.98989873, 197.98989873 >


The last set of steps is to compute the magnitude of the resultant.
r = magnitude = length of the vector
r = sqrt(x^2 + y^2) due to the Pythagorean theorem
r = sqrt( 357.98989873^2 + 197.98989873^2 )
r = <font color=red>409.09261493 newtons</font>
This answer is approximate. Round it however your teacher instructs.
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