Physics
posted by Matt on .
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?

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 
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

And your answer doesn't match the back of the book either.

The back of the book says 227 km/h at 162.4 degrees

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