Question 1073348
{{{drawing(350,250,-550,150,-100,400,
red(arrow(0,0,0,200)),locate(10,220,red(N)),
red(arrow(-176.78,176.78,-176.78,376.78)),locate(-166.78,397,red(N)),
circle(0,0,5),locate(-50,0,start),locate(10,25,S),
circle(-176.78,176.78,5),locate(-166,205,M),
circle(-501.24,364.28,5),locate(-495,395,E),
arrow(0,0,-176.78,176.78),
arrow(-176.78,176.78,-501.24,364.28),
green(arc(0,0,150,150,-90,225)),locate(80,30,green(315^o)),
green(triangle(-53.03,53.05,-66,53.05,-53.03,40)),
green(arc(-176.78,176.78,150,150,-90,210)),locate(-97,185,green(300^o)),
green(triangle(-241.73,214.28,-256.73,214.28,-241.73,200)),
locate(-170,100,250km),locate(-435,288,375km)
)}}}
 
I believe the westward distance SW (below) is what the problem asks for.
{{{drawing(350,250,-550,150,-100,400,
red(arrow(0,0,0,200)),locate(10,220,red(N)),
red(arrow(-176.78,176.78,-176.78,376.78)),locate(-166.78,397,red(N)),
circle(0,0,5),locate(-50,0,start),locate(10,25,S),
circle(-176.78,176.78,5),locate(-166,205,M),
circle(-501.24,364.28,5),locate(-495,395,E),
triangle(0,0,-176.78,176.78,-176.78,0),
triangle(-176.78,176.78,-501.24,364.28,-501.24,176.78),
rectangle(-176.78,176.78,-501.24,0),circle(-501.24,0,5),
locate(-510,0,W),circle(-176.78,0,5),circle(-501.54,176.78,5),
green(arc(0,0,150,150,180,225)),locate(-120,50,green(45^o)),
green(arc(-176.78,176.78,150,150,180,210)),locate(-300,230,green(30^o)),
locate(-95,120,250km),locate(-435,288,375km),
rectangle(-176.78,0,-157,20),rectangle(-501.54,176.78,-481,197),
locate(-180,0,U),locate(-520,200,A)
)}}}
(In the drawing above SUW and MA are East-West lines,
while UM and WAE area South-North lines).
To calculate the Westward distance SW,
we use right triangles SUM and MAE to calculate SU amd MA,
and then (since MA = UW) add the calculated distances to get
SW = SU + UW = SU + MA .
 
{{{SU/"250km"=cos(45^o)}}} --> {{{SU=(250km)*cos(45^o)}}}
Since {{{cos(45^o)=about0.707}}} ,
{{{SU=(250km)*0.707=about177km}}} is a good approximation.
{{{MA/"375 km"=cos(30^o)}}} --> {{{MA=(375km)*cos(30^o)}}}
Since {{{cos(30^o)=about0.866}}} ,
{{{MA=(375km)*0.866=about325km}}} is a good approximation.
So, a good approximation for SW is
{{{177km+325km=highlight(502km)}}}


NOTE:
The total distance SE is more difficult to find,
and it would be a typical pre-calculus problem in my school district,
because the most practical way to find SE
would be applying law of cosines to triangle SME,
knowing that the exterior angle at M measures {{{315^o-300^o=15^o}}} ,
and that makes the measure of its supplementary angle, SME, {{{180^o-15^o=165^o}}}.
For triangle SME, the law of cosine says that
{{{SE^2=SM^2+ME^2-2*(SM)*(ME)*cos(SME)}}} 
Substituting {{{SM=250km}}} , {{{ME=375km}}} ,
and {{{cos(SME)=cos(165^o)=-cos(15^o)=about-0.966}}} ,
However, that is not the only way to find SE,
so the task to find SE could be given to students in earlier mat courses,
as long as they know how ro solve right triangles using sine and cosine trigonometric ratios.