+5265 Need help FAST!!!
Disregard ssm as it means the answer is in something I do not have.
+129852 Writing the vector components thusly :
A = < 2cos 0° , 2sin 0° >
B = < 3.75 cos 90°, 3.75 sin 90° >
C = < 2.5 cos 180°. 2.5 sin 180° >
D = < 3 cos 270°, 3 sin 270° >
Sum the x components = 2cos 0° + 3.75cos 90° + 2.5cos 180° + 3cos 270° = 2 + 0 - 2.5 + 0 = -.5
Sum the y components = 2sin 0° + 3.75sin 90° + 2.5 sin 180° + 3 sin 270° = 0 + 3.75 + 0 - 3 = .75
Resultant vector = < x , y > = < -.5, .75 > → in the second quadrant
Resultant magnitude = sqrt ( x^2 + y^2 ) = sqrt [ (-.5)^2 + (.75)^2 ] ≈ . 90134 km
Direction = tan-1 (y/x) = tan-1 ( .75 / -.5) = tan-1 ( -3/2) = tan-1 (-1.5) ≈ -56.31°
However....this is a second quadrant angle so, with respect to due west, it is 56.31° north of due west
[i.e. 123.69° in standard position ]
