Question #94512
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.
1
Expert's answer
2019-09-16T09:48:44-0400

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°60\degree to 120°120\degree and eventually arrives to a town due east from the origin, we can deduct that it is initially 60°60\degree "east of south", whereas the latter angle 120°120\degree "east of south" would in other words mean (180120)°=60°(180-120)\degree = 60\degree "east of north".


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

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


Since then the aircraft flies at the same degree (60°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:

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


Result:

ΔsXZ=ΔsXY+ΔsYZ=2ΔsXY=230 km sin(60°)=303 km 52 km.\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}.


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