Answer to Question #94512 in Physics for April

Question

a plane moves from a town X to a town Y 30 km apart in the direction 60° and then changes direction moving to a point Z in the direction 120° if the point Z, is due east of X, find the distance of the point Z from point X.

Answer 1

In order to have the complete problem description, whenever a direction of flight (e.g. in degrees) is given, it should also be mentioned in reference to which axis/bearing the direction (i.e. angle) is.

If the aircraft reroutes from $60\degree$ to $120\degree$ and eventually arrives to a town due east from the origin, we can deduct that it is initially $60\degree$ "east of south", whereas the latter angle $120\degree$ "east of south" would in other words mean $(180-120)\degree = 60\degree$ "east of north".

With the first flight X to Y the displacement along "east" axis, in the east direction,

$\Delta s_{XY} = 30\text{ km }\cdot\sin(60\degree)$

Since then the aircraft flies at the same degree ($60\degree$) with respect to "east of north", it compensates for the same opposite displacement along the "north-south" axis, thus arriving at a town Z due east of X, therefore displacement Y to Z along the "east" axis is symetrically the same as from X to Y:

$\Delta s_{YZ} = \Delta s_{XY} = 30\text{ km }\cdot\sin(60\degree)$

Result:

$\Delta s_{XZ} = \Delta s_{XY} + \Delta s_{YZ} = 2\Delta s_{XY} = 2\cdot30\text{ km }\cdot\sin(60\degree) = 30\sqrt{3}\text{ km } \approx 52\text{ km}.$