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=-172.2-43.5= -216 km/h ....

Thank you. So much.

To determine whether the answer in the book is correct or not, let's go through the calculation steps together.

Given:
v = 190 km/h (velocity of the plane)
α = 25° (angle of the plane's direction)
u = 45 km/h (velocity of the wind)
β = 15° (angle of the wind's direction)

Step 1:
We need to find the x-component and y-component of the new velocity V(x) and V(y), respectively.

V(x) = v • cos(α) - u • sin(β)
= 190 km/h • cos(25°) - 45 km/h • sin(15°)
≈ 168.23 - 11.51
≈ 156.72 km/h

V(y) = v • sin(α) + u • sin(β)
= 190 km/h • sin(25°) + 45 km/h • sin(15°)
≈ 82.69 + 11.51
≈ 94.20 km/h

Step 2:
Now we can calculate the magnitude of the new velocity V using the Pythagorean theorem.

V = √(V(x)² + V(y)²)
= √((156.72 km/h)² + (94.20 km/h)²)
= √(24599.58 + 8872.44)
= √(33472.02)
≈ 183.04 km/h

Therefore, based on the calculations, the correct answer for the magnitude of the plane's new velocity with respect to the ground in standard position is approximately 183.04 km/h.

Since the book's answer states that V = 227 km/h, there may be a mistake in the book's solution. It is advisable to double-check the calculation steps and ensure that the angles and velocities are entered correctly.