1. An airplane is flying with the velocity of 459km/h [E40S], however its ground velocity is 498km/h [E24S]. Determine the velocity of the blowing wind. (Use the vector subtraction diagram and geometry to solve this question).

I drew a diagram and tried solving this question by using components method.
so basically...

Vwg = 498km/h [E24S]-459km/h[E40S]
Vwg = 498km/h [E24S]+ 459km/h[W40N]
Vwg = 498km/h*cos24 +298km/h*sin24 + 459*cos40 + 459*sin40
Vwg = 455km/h[E]+203km/h[S]+351km/h[W] + 295km/h[N]
Vwg = 104km/h[E]+ 92km/h[N]
Vwg = 138.86km/h[E41.5N]

so i got 138.86km/h
1. i don't know if i am correct
2. I don't understand how to use geometry to solve the question. i drew the diagram but i don't know if that is good. I also don't get how to do vector subtraction.

Please help
thank you

All angles are CCW.

459km/h[140o] + Vw = 498km/[156o]
Vw = 498[156] - 459[140]

X = 498*Cos156 - 459*Cos140 =-103.3 km/h
Y = 498*sin156 - 459*sin140 = -92.48km/h

Tan Ar = Y/X = -92.48/-103.3 = 0.89530
Ar = 41.84o = Reference angle.
A = 41.84 + 180 = 221.84o CCW = 41.84o
S. of W.(W41.84S).

Vw = X/Cos A = -103.3/Cos221.84 = 138.7
km/h[W41.84S].

Note: Since X and Y are both negative,
the resulting velocity is in the 3rd
Quadrant.

To determine the velocity of the blowing wind, you need to subtract the velocity of the airplane from the ground velocity of the airplane. Here's a step-by-step explanation:

1. Start by drawing a diagram to represent the situation. The diagram should include the vectors for the ground velocity (498 km/h [E24S]) and the velocity of the airplane (459 km/h [E40S]).

2. To perform vector subtraction, you need to change the direction of the airplane's velocity vector. Since the direction is given as [E40S], you can change it to [W40N] by reversing both the direction and the angle, resulting in a velocity vector of 459 km/h [W40N].

3. Add the two vectors together by adding the corresponding components separately. For the components in the east-west direction, add 498 km/h (from the ground velocity vector) and 459 km/h (from the airplane's velocity vector) to get a total of 957 km/h [E]. For the components in the north-south direction, subtract the south component of the ground velocity (24 degrees) from the north component of the airplane's velocity (40 degrees) to get a total of 92 km/h [N].

4. The resultant vector of the wind velocity is the combination of the east-west and north-south components. Based on your calculations, you obtained a wind velocity of 138.86 km/h in the direction [E41.5N].

Now, to address your questions:

1. Based on your calculation, you obtained a wind velocity of 138.86 km/h [E41.5N]. It means that the wind is blowing at a speed of 138.86 km/h towards the east-northeast direction.

2. Using geometry to solve this question involves breaking down the vectors into their x (east-west) and y (north-south) components and applying basic trigonometry to find the magnitudes of the components. From your calculations, it seems like you did this correctly, using the cosine and sine functions to find the x and y components respectively of both vectors and then adding them together.

Overall, your approach and calculations seem correct, and you have arrived at the correct answer for the wind velocity. Just make sure to double-check your calculations and consider the sign conventions for the components.