An airplane has an airspeed of 650 km/h bearing 15 degrees north of west. The wind velocity of is 19 km/h in the direction of 30 degrees east of south. Find the resultant vector representing the path of the plane relative to the ground. What is the ground speed of the plane? What is its direction?

Note: I know how to solve just need help with finding the correct angles for problem.

Carefully draw the two vectors, placing the start of the wind vector at the end of the plane's vector. If you draw in some dotted axes, you can see that the angle between the vectors (geometric, not algebraic) is 90-30-15 = 45°

The ground speed of the plane is found using the law of cosines, since you now have SAS.

To get the direction, you need to convert to x-y components, add the two vectors and find the angle whose tangent is the resultant y/x.

To find the resultant vector representing the path of the plane relative to the ground, we need to find the vector sum of the airplane's airspeed and the wind velocity.

Step 1: Break down the airspeed and wind velocity into their horizontal and vertical components.

The airspeed of 650 km/h bearing 15 degrees north of west can be broken down into horizontal and vertical components as follows:

- Airspeed horizontal component: 650 km/h * cos(15°)
- Airspeed vertical component: 650 km/h * sin(15°)

The wind velocity of 19 km/h in the direction of 30 degrees east of south can be broken down into horizontal and vertical components as follows:

- Wind velocity horizontal component: 19 km/h * cos(30°)
- Wind velocity vertical component: -19 km/h * sin(30°) [negative sign because south is in the opposite direction of the positive vertical direction]

Step 2: Add the horizontal and vertical components separately to determine the resultant vector.

- Resultant horizontal component: Airspeed horizontal component + Wind velocity horizontal component
- Resultant vertical component: Airspeed vertical component + Wind velocity vertical component

Step 3: Calculate the magnitude and direction of the resultant vector.

The magnitude of the resultant vector (ground speed of the plane) can be found using the Pythagorean theorem:

Ground speed = sqrt(Resultant horizontal component^2 + Resultant vertical component^2)

The direction of the resultant vector can be found using:

Direction = atan(Resultant vertical component / Resultant horizontal component) + 180°

Now, you can substitute the values into these formulas to find the ground speed and direction of the plane.

To solve this problem, we need to break it down into two steps: finding the resultant vector representing the path of the plane relative to the ground, and then finding the ground speed and direction of the plane.

Step 1: Finding the resultant vector:

To find the resultant vector, we need to add the vectors representing the airplane's airspeed and the wind velocity. Let's assume that the positive x-axis points east, and the positive y-axis points north.

The airspeed vector (A) has a magnitude of 650 km/h and is directed 15 degrees north of west. To represent it in component form, we need to find its x and y components.

- The x-component of A is A_x = |A| * cos θ = 650 km/h * cos(15°).
- The y-component of A is A_y = |A| * sin θ = 650 km/h * sin(15°).

The wind velocity vector (W) has a magnitude of 19 km/h and is directed 30 degrees east of south. To represent it in component form, we need to find its x and y components.

- The x-component of W is W_x = |W| * sin θ = 19 km/h * sin(30°).
- The y-component of W is W_y = -|W| * cos θ = -19 km/h * cos(30°). (Note: The negative sign is because south is in the negative y direction.)

Now, we can find the resultant vector (R) by adding the x and y components of A and W:

R_x = A_x + W_x
R_y = A_y + W_y

Step 2: Finding the ground speed and direction:

To find the ground speed and direction of the plane, we need to find the magnitude and direction of the resultant vector.

The magnitude of the resultant vector (|R|) is given by the Pythagorean theorem:

|R| = √(R_x^2 + R_y^2)

The direction of the resultant vector (θ_R) can be found using the inverse tangent function:

θ_R = atan2(R_y, R_x)

Now, you can calculate the resultant vector (path of the plane relative to the ground), its ground speed |R|, and its direction θ_R using the formulae mentioned above.