After flying for 15 min in a wind blowing 42 km/h at an angle of 28° south of east, an airplane pilot is over a town that is 42 km due north of the starting point. What is the speed of the airplane relative to the air?

To find the speed of the airplane relative to the air, we need to break down the velocity vectors into their components.

1. The wind is blowing at a speed of 42 km/h at an angle of 28° south of east. The components of the wind's velocity can be found using trigonometry:
- The horizontal component of the wind velocity is given by: Vx = wind speed * cos(angle)
Vx = 42 km/h * cos(28°)
Vx ≈ 37.9 km/h
- The vertical component of the wind velocity is given by: Vy = wind speed * sin(angle)
Vy = 42 km/h * sin(28°)
Vy ≈ 19.0 km/h

2. After flying for 15 minutes (0.25 hours) in the wind, the airplane has traveled a distance of:
Distance = speed * time
Distance = unknown * 0.25 hours

3. The airplane is over a town that is 42 km due north of the starting point. Therefore, the vertical distance the airplane has traveled is 42 km.

4. From the given information, the vertical component of the airplane's velocity relative to the air is equal to the vertical distance traveled in the time it took to get there:
Vy = 42 km / 0.25 hours
= 168 km/h

5. Using Pythagorean theorem, we can find the speed of the airplane relative to the air:
Speed = √(Vx^2 + Vy^2)
Speed = √((37.9 km/h)^2 + (168 km/h)^2)
Speed ≈ 172 km/h

Therefore, the speed of the airplane relative to the air is approximately 172 km/h.

To find the speed of the airplane relative to the air, we need to break down the velocity vectors into their components.

Let's assume the speed of the airplane relative to the air is "v" km/h, and the angle the airplane is flying relative to the east direction is "θ".

Now, let's break down the velocity of the airplane relative to the air into its components.

The eastward component of the velocity can be calculated as:
Ve = v * cos(θ)

The southward component of the velocity can be calculated as:
Vs = -v * sin(θ)

In this case, we know that the southward component of the wind velocity is 42 km/h, and the angle of the wind relative to the east direction is 28°.

Let's break down the wind velocity into its components.

The eastward component of the wind velocity can be calculated as:
We = 42 km/h * cos(28°)

The southward component of the wind velocity can be calculated as:
Ws = -42 km/h * sin(28°)

Since the airplane is flying directly north from its starting point, the eastward components of the airplane's velocity and the wind's velocity cancel each other out.

Ve + We = 0

Therefore, we have:
v * cos(θ) + 42 km/h * cos(28°) = 0

Now, let's use the given information that after 15 minutes of flying, the airplane is over a town that is 42 km due north of the starting point.

The distance traveled northward by the airplane can be calculated as:
Distance northward = (v * sin(θ) + 42 km/h * sin(28°)) * (15 min / 60 min/h)

Since the airplane is 42 km due north of the starting point, we have:
(v * sin(θ) + 42 km/h * sin(28°)) * (15 min / 60 min/h) = 42 km

Now, we can solve these two equations simultaneously to find the speed of the airplane relative to the air (v).

Solving these equations may involve some trigonometric calculations. Let's calculate the values and find the speed of the airplane relative to the air.