Posted by Tabby on Monday, April 23, 2012 at 8:23pm.
probably easier for you to do this on x, y axis system
First find x and y components of first force (15deg above x axis)
F1x = 35 cos 15
F1y = 35 sin 15
now the second force(90-55) = 35 below -x axis)
F2x = -75 cos 35
F2y = -75 sin 35
now add components
Fx = 35 cos 15 -75 cos 35 = -27.6
Fy = 35 sin 15 -75 sin 35 = -34.0
F^2 = Fx^2+Fy^2
so F = 43.7 in magnitude
tan theta = Fy/Fx = -43/-27.6
in quadrant 3 because both are negative
so theta = 57.3 deg below -x axis
or 57.3 deg south of west
or 90-57.3 = 32.7 deg west of south
Ah, thank you so much! You're a lifesaver! Thank you and God bless you for taking the time to help!
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