An airplane flies at airspeed (relative to the air) of 430 km/h . The pilot wishes to fly due North (relative to the ground) but there is a 43 km/h wind blowing Southwest (direction 225◦).

In what direction should the pilot head the plane (measured clockwise from North)?
Answer in units of degrees.

is 41 degrees east of north the exact degree number??? because a computer program wants the exact number and i cant round over or under 1%...

To determine the direction the pilot should head the plane, we need to find the resultant velocity vector. We can use vector addition to calculate this.

1. First, we need to break down the wind velocity into its horizontal and vertical components relative to the ground.

Given:
- Wind speed: 43 km/h
- Wind direction: 225° (southwest)

To find the horizontal component of the wind velocity, we can use the cosine function:
Horizontal component = Wind speed × cos(Wind direction)

Horizontal component = 43 km/h × cos(225°)

Horizontal component = 43 km/h × (-0.7071) (rounded to four decimal places)

Horizontal component = -30.3763 km/h (rounded to four decimal places)

To find the vertical component of the wind velocity, we can use the sine function:
Vertical component = Wind speed × sin(Wind direction)

Vertical component = 43 km/h × sin(225°)

Vertical component = 43 km/h × (-0.7071) (rounded to four decimal places)

Vertical component = -30.3763 km/h (rounded to four decimal places)

So, the wind velocity is decomposed into a horizontal component of -30.3763 km/h and a vertical component of -30.3763 km/h.

2. Now, we can find the resultant velocity vector by adding the airspeed vector (430 km/h) to the wind velocity vector.

Let's call the resultant velocity vector R.
R = Airspeed + Wind velocity

To add the vectors, we need to find the horizontal and vertical components separately.

Horizontal component of R = Airspeed (horizontal) + Wind velocity (horizontal)
Vertical component of R = Airspeed (vertical) + Wind velocity (vertical)

Horizontal component of R = 430 km/h + (-30.3763 km/h)

Horizontal component of R = 399.6237 km/h (rounded to four decimal places)

Vertical component of R = 0 km/h + (-30.3763 km/h)

Vertical component of R = -30.3763 km/h (rounded to four decimal places)

The resultant velocity vector R has a horizontal component of 399.6237 km/h and a vertical component of -30.3763 km/h.

3. Finally, we can find the direction of the resultant velocity vector R by using the inverse tangent (arctan) function.

Direction angle = arctan(Vertical component of R / Horizontal component of R)

Direction angle = arctan((-30.3763 km/h) / (399.6237 km/h))

Direction angle ≈ -04.34° (rounded to two decimal places)

Since the question asks for the direction measured clockwise from North, we can conclude that the pilot should head the plane in the direction of approximately 4.34° south of due east (measured clockwise).

To determine the direction the pilot should head the plane, we need to calculate the resultant velocity vector by adding the velocity of the airplane to the velocity of the wind.

Let's break down the airspeed and wind velocity into their x and y components.

The airspeed of the plane can be represented as:
V_plane_x = airspeed * cos(0◦) = 430 km/h * cos(0◦) = 430 km/h

V_plane_y = airspeed * sin(0◦) = 430 km/h * sin(0◦) = 0 km/h

The wind velocity can be represented as:
V_wind_x = wind speed * cos(225◦) = 43 km/h * cos(225◦) ≈ -30.404 km/h

V_wind_y = wind speed * sin(225◦) = 43 km/h * sin(225◦) ≈ -30.404 km/h

To find the resultant velocity, we can add the x and y components of the airspeed and wind velocity:

V_resultant_x = V_plane_x + V_wind_x = 430 km/h + (-30.404 km/h) = 399.596 km/h

V_resultant_y = V_plane_y + V_wind_y = 0 km/h + (-30.404 km/h) = -30.404 km/h

Next, we can calculate the direction of the resultant velocity (measured clockwise from North). We can use the inverse tangent function to find the angle:

θ = arctan(V_resultant_y / V_resultant_x)

θ = arctan(-30.404 km/h / 399.596 km/h)

Using a calculator, we can find that θ ≈ -4.34 degrees.

Since we are measuring the angle clockwise from North, the direction the pilot should head the plane is 360◦ - (-4.34◦) = 364.34◦.

Therefore, the pilot should head the plane in a direction approximately 364.34◦ clockwise from North.

43/√2 = 30.4, so the wind's vector is (-30.4,-30.4)

If the plane flies on heading θ,

(-30.4,-30.4)+(430cos(θ-45°),430sin(θ-45°)) = (0,y)

so, we have -30.4 + 430cos(θ-45) = 0
cos(θ-45) = 30.4/430 = 4.05

so, the plane flies N41°E