Answer:
The direction will be [tex]84.86^\circ[/tex] and the distance 250.75km.
Explanation:
Let's say A is the displacement vector which represents the first 170km and B the one for the next 230km. Then the components of these vector will be:
[tex]A_x=170cos(68^{\circ})\\ A_y=170sin(68^\circ)\\\\B_x=230cos(-36^\circ)\\B_y=230sin(-36^\circ)[/tex]
The vector which point from the origin to the final position of the plane will be R=A+B. We sum components on x and y independetly (vector property):
[tex]R_x=A_x+B_x=170cos(68^{\circ})+230cos(-36^\circ)=63.68km+186.07km=249.75km[/tex]
[tex]R_y=A_y+B_y=170sin(68^\circ)+230sin(-36^\circ)=157.62km-135.19km=22.43km[/tex]
If [tex]\theta[/tex] is the direction of R then:
[tex]tan(\theta )=\frac{R_x}{R_y}[/tex] ⇒ [tex]\theta = arctan(\frac{R_x}{R_y})[/tex] ⇒ [tex]\theta = 84.86^\circ[/tex].
The distance will be given by the magnitud of the vector R:
[tex]R=\sqrt{R_x^2 + R_y^2}[/tex] ⇒ [tex]R=\sqrt{R_x^2 + R_y^2} = 250.75[/tex].
