Law of Cosine and Sine for two Forces: Based on figures in simulation: R = ? : 30 Question: Can please FB = 200 N; 180 B = 126.2 someone explain where FA = 100 N; 30 did 126.2 came from? Or how to to get Law of Cosine: 126 . 2? R = F + F - 2(F . F.)cos(Y) R' = (100N) + (200N) - 2(100N . 200N) cos(309) R = 15 358.98385 N R = V15 358. 98385 N" = 123. 9314 N ~ 123.93 N Law of Sine: R (R) sin (B) sin(B) sin(Y) = y = sin FB Y = sin 1(123.9314 N) sin (126.2 ) = 30. 0 (200 N) 0.= y + B = 30.0 + 126.2 = 156.2 Therefore, the resultant vector have a magnitude of 123.93 N and direction of 156.2.- 3 12.4 156.2 -11.3 20 Idi a 10.0 171 2 12.4 -20 le1 - 20.0