An airplane needs to head due north, but there is a wind blowing from the southwest at 85 km/hr. The plane flies with an airspeed of 550 km/hr. To end up flying due north, how many degrees west of north will the pilot need to fly the plane? Round to the nearest tenth.
To determine how many degrees west of north the pilot needs to fly the plane, we can use trigonometry.
Let's first find the angle between the direction of the plane's heading and north. This angle is formed by the north direction and the direction of the plane's velocity vector, which is the sum of the airspeed vector and the wind vector.
Let's assume the angle between the north direction and the direction of the plane's velocity vector (resultant vector) is represented by θ.
Using the cosine function, we can calculate θ:
cos(θ) = (horizontal component of the resultant vector) / (resultant vector)
The horizontal component of the resultant vector is the component in the direction west of north. In this case, it would represent the component due to the wind. We can calculate this component using the wind speed and the given angle of the wind direction.
Let's assume the angle of the wind direction (southwest) from the north direction is represented by α.
The horizontal component of the resultant vector can be found using the formula:
(horizontal component) = (wind speed) * cos(α)
In this case, the wind speed is 85 km/hr.
To find α, we need to find the angle the wind makes with the west direction, which is 90 degrees minus the angle between north and the wind direction. So:
α = 90 - (90 - 45) = 45 degrees
Substituting the values into the formula, we have:
(horizontal component) = 85 * cos(45)
To calculate cos(45), we need to convert the angle to radians:
45 degrees = (45 * π) / 180 radians
(cosine function uses radians)
(horizontal component) = 85 * cos((45 * π) / 180)
Now, we can calculate θ. Rearranging the cosine function equation:
cos(θ) = (resultant vector - horizontal component) / resultant vector
The resultant vector is given by the airspeed of the plane, which is 550 km/hr.
Using the values, we have:
θ = arccos((550 - (85 * cos((45 * π) / 180))) / 550)
Calculating this using a calculator, we find:
θ ≈ 11.9 degrees
Therefore, rounded to the nearest tenth, the pilot will need to fly approximately 11.9 degrees west of north to end up flying due north.
To determine how many degrees west of north the pilot needs to fly the plane, we can break down the vectors involved.
Let's consider the airplane's airspeed and the wind as vectors. The airplane's airspeed vector points due north, while the wind vector points from the southwest.
First, we need to find the resultant vector of the airplane's airspeed and the wind vector. We can use vector addition to find this resultant vector.
Next, we will find the angle between the resultant vector and the north direction. This angle will represent the number of degrees west of north that the pilot needs to fly the plane.
To start, we can draw a diagram to visualize the vectors involved. Draw a north-oriented line and label it "North" at the top of the page. Then, draw a line pointing southwest, which represents the wind vector. Label this line "Wind."
Next, draw a line from the North line representing the airplane's airspeed vector. This line should be tilted slightly to the right to indicate that it is not directly north. Label this line "Airspeed."
Now, we can measure the magnitude of each vector. The magnitude of the wind vector is 85 km/hr (given in the problem). The magnitude of the airplane's airspeed vector is 550 km/hr (also given in the problem).
Using vector addition, draw a line from the end of the Airspeed line to the end of the Wind line. This line represents the resultant vector.
To find the magnitude of the resultant vector, we can use the Pythagorean theorem. The magnitude of the resultant vector can be calculated as the square root of the sum of the squares of the magnitudes of the airspeed and wind vectors.
Magnitude of resultant vector = √(magnitude of airspeed vector)^2 + (magnitude of wind vector)^2
Magnitude of resultant vector = √(550^2 + 85^2)
Magnitude of resultant vector ≈ √(302,500 + 7,225)
Magnitude of resultant vector ≈ √(309,725)
Magnitude of resultant vector ≈ 556.3 km/hr (rounded to the nearest tenth)
Now that we know the magnitude of the resultant vector, we can find the angle it forms with the north direction using trigonometry. Specifically, we can use the inverse tangent function (tan^-1).
Angle = tan^-1 (magnitude of wind vector / magnitude of airspeed vector)
Angle = tan^-1 (85 / 550)
Angle ≈ tan^-1 (0.15455)
Angle ≈ 8.89 degrees
Therefore, the pilot needs to fly approximately 8.9 degrees west of north to cancel out the effect of the wind and end up flying due north (rounded to the nearest tenth).
85 kph * sin
(45º) = 550 kph * sin(Θ)
Θ is the angle west of north