Question

An airplane trip involves three legs with two stop overs. The first leg is due east for 620km, the second leg is due is south east (45°) for 440km and the third leg is at 53° south of west for 550km. Determine the airplanes resultant displacement (vector) using the graphical technique. Scale is 1:100km

Answer 1

$$D_x=620+440\cos{45}-550\cos{53}=600.13\ km\\ D_y=-440\sin{45}-550\sin{53}=-750.38\ km\\ D=\sqrt{600.13^2+750.38^2}=960.85\ km$$

$$\theta=\arctan{\frac{750.38}{600.13}}=51.3\degree \text{ S of E}$$