[S2°W] ----> 268°
[N31°W] ---> 121°
Resultant vector
= 9(cos268, sin268) + 11(cos121,sin121)
= (-.03141, -8.9945) + (-5.6654, 9.4288)
= (-5.9795, 0.4343)
magnitude = 5.9953
direction = 175.85°
or
make a sketch, on mine I can use the cosine law to get
R^2 = 9^2 + 11^2 - 2(9)(11)cos33°
= 35.9432...
R = √35.9432... = 5.9953 , same as before.
I will leave it up to you to find the direction angle using the sine law
Add the following vectors using trigonometry (i.e. cosine and sine laws).
9 N on a heading of [S2°W] and 11 N on a heading of [N31°W].
2 answers
F=r = 9{182o] + 11[329o].
X = 9*sin182 + 11*sin329 = -5.98 N.
Y = 9*cos182 + 11*Cos329 = 0.4343 N.
Fr = sqrt(X^2 + Y^2) = 5.996 N. = Resultant force.
TanA = X/Y, A = -85.9o = 85.9o W. of N. = 274.1 CW from +Y-axis = 175.9o CCW from +X-axis
X = 9*sin182 + 11*sin329 = -5.98 N.
Y = 9*cos182 + 11*Cos329 = 0.4343 N.
Fr = sqrt(X^2 + Y^2) = 5.996 N. = Resultant force.
TanA = X/Y, A = -85.9o = 85.9o W. of N. = 274.1 CW from +Y-axis = 175.9o CCW from +X-axis