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Sagot :
probaly you meant 'what bearing should be taken'
so use pythagorean theorem
c=distance from position to south port
b=45
a=30
a^2=B^2=c^2
30^2+45^2=c^2
900+2025=c^2
2925=c^2
square root of both sides
15 times √13
aprox
54.08
54.1
opposite of east=west
opposite of south=north
bearing=
54.1 miles north-west
so use pythagorean theorem
c=distance from position to south port
b=45
a=30
a^2=B^2=c^2
30^2+45^2=c^2
900+2025=c^2
2925=c^2
square root of both sides
15 times √13
aprox
54.08
54.1
opposite of east=west
opposite of south=north
bearing=
54.1 miles north-west
Well
East
45 miles > > > >
< v
< v 30 miles South
North < v
West < Δ= ship
A right angle is formed, so we use Pythagoras.
[tex]45 ^{2} +30 ^{2}=2925 [/tex]
[tex] \sqrt{2925}=54.08 [/tex]
Therefore, the route should be 54 miles North West
East
45 miles > > > >
< v
< v 30 miles South
North < v
West < Δ= ship
A right angle is formed, so we use Pythagoras.
[tex]45 ^{2} +30 ^{2}=2925 [/tex]
[tex] \sqrt{2925}=54.08 [/tex]
Therefore, the route should be 54 miles North West
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