Asked by Gigi
A ship travels due north for 25 miles then changes course to a bearing of 300°. It travels on this path for 29 miles. How far is the ship from where it started?
Answers
Answered by
MathMate
Sum the east (x-) and north (y-) components.
x=∑D cos(θ)
=25*cos(90°)+29 cos(90-300°)
=0+29cos(-210)
=-(29/2)√3
y=∑D sin(θ)
=25*sin(90°)+29sin(90-300°)
=25+29sin(-210°)
=25+29sin(30°)
=25+14.5
=39.5
Distance
=sqrt(x^2+y^2)
=...
x=∑D cos(θ)
=25*cos(90°)+29 cos(90-300°)
=0+29cos(-210)
=-(29/2)√3
y=∑D sin(θ)
=25*sin(90°)+29sin(90-300°)
=25+29sin(-210°)
=25+29sin(30°)
=25+14.5
=39.5
Distance
=sqrt(x^2+y^2)
=...
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