Asked by Travis
Airplane staring from airport a flies 300 km east then 395 km 26.5 degrees west of north, and then 150 km north to arrive at airport b. what angle is the straight flight from a to b north of east
and how far
and how far
Answers
Answered by
Elena
Make the route plan
s₁= 300 =x₁+x₂
s₂ =395, α =26.5⁰
s₃ = 150
x₃ connects the point of s₂ and s₃ intersection and the point on s₁
which divides s₁ by x₁ and x₂ (x₃ ⊥ s₁)
s₄=? β =?
x₂ = s₂•sin α = 395 •sin 26.5 =176.25 km
x₃ =s₂•cos α = 395 •cos26.5 =353.5 km
s₃+ x₃ = 353.5+150 = 503.5 km
x₁ =s₁-x₂ =300-176.25 = 123.75 km
s₄ =sqrt{ (s₃+ x₃)²+x₁²} =
=sqrt(503.5²+123.75²) =518.5 km.
tan β = (s₃+ x₃)/x₁ =503.5/123.75 = 40.69
β = 88.6⁰
s₁= 300 =x₁+x₂
s₂ =395, α =26.5⁰
s₃ = 150
x₃ connects the point of s₂ and s₃ intersection and the point on s₁
which divides s₁ by x₁ and x₂ (x₃ ⊥ s₁)
s₄=? β =?
x₂ = s₂•sin α = 395 •sin 26.5 =176.25 km
x₃ =s₂•cos α = 395 •cos26.5 =353.5 km
s₃+ x₃ = 353.5+150 = 503.5 km
x₁ =s₁-x₂ =300-176.25 = 123.75 km
s₄ =sqrt{ (s₃+ x₃)²+x₁²} =
=sqrt(503.5²+123.75²) =518.5 km.
tan β = (s₃+ x₃)/x₁ =503.5/123.75 = 40.69
β = 88.6⁰
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