In unit vector form
Fx=20cos30−40cos60−25+12cos45=−19.194NFy=20sin30−40sin60−42+12sin45=−5.844NF=Fx2+Fy2F=(−19.194)2+(−5.844)2F=20.064Nα=tan−1(−5.844−19.194)α=16.9340F_x = 20 cos 30 - 40 cos 60 -25 +12 cos 45 = - 19.194 N\\ F_y = 20 sin 30 - 40 sin 60 -42 +12 sin 45 = - 5.844 N\\ F= \sqrt{F_x^2+F_y^2}\\ F= \sqrt{(-19.194)^2+(-5.844)^2}\\ F= 20.064 N\\ \alpha = \tan^{-1}(\frac{-5.844}{-19.194})\\ \alpha = 16.934^0\\Fx=20cos30−40cos60−25+12cos45=−19.194NFy=20sin30−40sin60−42+12sin45=−5.844NF=Fx2+Fy2F=(−19.194)2+(−5.844)2F=20.064Nα=tan−1(−19.194−5.844)α=16.9340
Therefore, the angle is 1800+16.9340 = 196.9340
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