Asked by Matt

A plane is flying 25 degrees north of west at 190 km/h and encounters a wind from 15 degrees north of east at 45 km/h. What is the planes new velocity with respect to the ground in standard position?
Answered by Elena
v=190 km/h, α= 25°,
u=45 km/h, β =15°.

V(x) =v•cos α - u•sin β =…
V(y)= v•sinα+u•sinβ=…
V=sqrt{V(x)²+V(y)²}=...

Sine Law:
V/sin(α+ β) =u/sinγ

Solve for γ.

V directed (α+γ)° north of west
Answered by Matt
lol can you solve this for me? We have done this multiple times (we think) and we are getting a different answer than that's in the back of the book
Answered by Matt
And your answer doesn't match the back of the book either.
Answered by Matt
The back of the book says 227 km/h at 162.4 degrees
Answered by Elena
V(x) = - v•cos α +u•cosβ = - 190cos25 + 45cos15=172.2+43.5= -128.7 km/h
V(y)= v•sinα+u•sinβ= 190sin25 + 45sin15= 80.3 + 11.6 = 91.9 km/h
V=sqrt{V(x)²+V(y)²}= 158.1 km/h
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