Two forces of magnitude 8N and 12N are acting on a body at angle 30 degree to each other.Find the resultant and direction on of the force of the body
Using basic trig, make a sketch and see that
R^2 = 8^2 + 12^2 - 2(8)(12)cos150
= 374.2768...
R = appr 19.35 N
by sine law:
sin150/19.35 = sinθ/12
sinθ = .310...
θ = 18.1° <---- between the 8 N and the resultant
by vectors
vector R = (12cos0, 12sin0) + (8cos30, 8sin30)
= (12, 0) + (4√3, 4)
= (12+4√3, 4)
|R| = √( (12+4√3)^2 + 16) = appr 19.346... , same as before
tan(angle) = 4/(12+4√3) = .2113..
angle = appr 11.932 <--- angle between 12 N and R
note that 11.9 + 18.1 = 30°