Question 43423
{{{drawing(400,400,-350,350,-350,350,
arrow(0,0,-285,0),
arrow(-285,0,-285,320),
arrow(0,0,-285,320),
locate(-275,150,320),
locate(-150,-5,285),
locate(0,-2,A),
locate(-285,-2,B),
locate(-275,335,C),
blue(arrow(0,-175,50,-175),arrow(0,-175,-50,-175),arrow(0,-175,0,-125),arrow(0,-175,0,-225)),
locate(55,-175,E),
locate(-55,-175,W),
locate(0,-115,N),
locate(0,-228,S),
locate(-40,35,X^o)
)}}}


Apply Pythagorus theorem.
{{{AC^2 = AB^2 + BC^2}}}
or {{{AC^2 = 285^2 + 320^2}}}
or {{{AC^2 = 81225 + 102400}}}
or {{{AC^2 = 183625}}}
or {{{AC = sqrt(183625)}}}
or {{{AC = 428.515}}}


Again, angle between AB and AC is 
{{{tan(X) = BC/AB}}}
or {{{tan(x) = 320/285 = 1.1228}}}
or X = tan inverse (1.1228) = {{{48.31^o}}}


So, the distance of the plane from its starting point is 428.515 km and the angle X is {{{48.31^o}}}.