Question

 Vector has magnitude of 20 km with quadrant bearing S 30° W and vector has magnitude of 35 km with quadrant bearing N 67°W. 

Answer 1

$$Solution$$

The angle between y-axis and $12N$ Vectors is ; Therefore

$\implies\ 360^\circ-\ 280^\circ\\ =80^\circ\\ Angle\ between\ vectors\\ \implies\theta=80+65\\ =\theta=145^\circ\\ Then\ using\ a\ parallelogram\\ Angle\ between\ vector\ \implies\ 180-\theta\\ \implies180-145\\ \implies\ 35^\circ\ Now,\ find\ the\ Resultant\ R\\ =\sqrt{(12)^2+(40)^2+2*12*40*Cos(35^\circ)}\\ =\sqrt{2530.38596}\\ =50.30\\ Direction(Use\ Sine)\\ =\frac{Sin(35^\circ)}{Sin\ \theta}=\frac{R}{40}\\ Sin\ \theta=\frac{40*Sin(35^\circ)}{50.30^\circ}\\ \theta=27.14^\circ$