PHYSICS
posted by Matt, AGAIN on .
This is the same question as last time.
"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?"
Elena kindly gave me this afterward
"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}=... "
Using this formula and inputting the number accordingly, it have me an answer of V = 171.4
The answer in the back of the book has an answer of V = 227
Is it possible the book answer is wrong?

I've corrected my mistake in the 1st equation
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
I believe that the answer from your book will be if
"...a wind from 15 degrees north of WEST at 45 km/h. "
The 1st equation will be
V(x) =  v•cos α  u•cosβ =  190cos25  45cos15=172.243.5= 216 km/h .... 
Thank you. So much.