Asked by Julio
A plane leaves Seattle, flies 84.0{\rm mi} at 21.0^\circ north of east, and then changes direction to 52.0^\circ south of east. After flying at 120{\rm mi} in this new direction, the pilot must make an emergency landing on a field. The Seattle airport facility dispatches a rescue crew.
In what direction should the crew fly to go directly to the field? Use components to solve this problem.
How far should the crew fly to go directly to the field? Use components to solve this problem.
In what direction should the crew fly to go directly to the field? Use components to solve this problem.
How far should the crew fly to go directly to the field? Use components to solve this problem.
Answers
Answered by
Henry
D = 84mi @ 21Deg. + 120mi @ -52Deg.
X = 84*cos21 + 120*cos(-52) = 152.3 Mi.
Y = 84*sin21 + 120*sin(-52) = -64.5 Mi.
tanA = Y/X = -64.5 / 152.3 = -.042343.
A = -22.9 Deg. = 22.9 Deg. S. of East.
D=X/cosA = 152.3 / cos(-22.9)=165.3 Mi.
= Dist.
X = 84*cos21 + 120*cos(-52) = 152.3 Mi.
Y = 84*sin21 + 120*sin(-52) = -64.5 Mi.
tanA = Y/X = -64.5 / 152.3 = -.042343.
A = -22.9 Deg. = 22.9 Deg. S. of East.
D=X/cosA = 152.3 / cos(-22.9)=165.3 Mi.
= Dist.
Answered by
Elena
a=84 mi, b=120 mi c=?
a•cos21+b•cos 55 = c•cosθ
b•sin55-a•sin21= c•sinθ
c•cosθ =84• •cos21+120•cos 55 =146.5 ...(1)
c•sinθ = 120•sin55-84•sin21 = 68.16 ...(2)
Divide (2) by (1)
tanθ =68.16/146.5 = 0.465
θ =arctan0.465 = 24.95º (south of east)
From (1)
c=146.5/cos24.95 = 161.6 mi
a•cos21+b•cos 55 = c•cosθ
b•sin55-a•sin21= c•sinθ
c•cosθ =84• •cos21+120•cos 55 =146.5 ...(1)
c•sinθ = 120•sin55-84•sin21 = 68.16 ...(2)
Divide (2) by (1)
tanθ =68.16/146.5 = 0.465
θ =arctan0.465 = 24.95º (south of east)
From (1)
c=146.5/cos24.95 = 161.6 mi
Answered by
Elena
My mistake: I've used 55 instead of 52
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