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To solve this problem, let's break down the situation into two parts: finding the angle off course and calculating the plane's ground speed.

### 1. Given Information:
- **Plane's airspeed** (relative to the air): \( 600 \, \text{km/hr} \) heading north.
- **Wind speed**: \( 50 \, \text{km/hr} \) blowing from the southwest. Since "southwest" means at a 45° angle to both the south and west, this wind vector has components in both the west and north directions.

### 2. Wind Components:
Since the wind is blowing from the southwest, its components can be found as:
\[
\text{Wind component in the west direction} = 50 \cos(45^\circ) = 50 \times \frac{\sqrt{2}}{2} = 35.36 \, \text{km/hr}
\]
\[
\text{Wind component in the north direction} = 50 \sin(45^\circ) = 50 \times \frac{\sqrt{2}}{2} = 35.36 \, \text{km/hr}
\]

### 3. Resultant Velocity Components:
The plane's northward velocity and the wind's northward component will combine, while the wind's westward component will push the plane off course.

- **Northward velocity** (combined): \( 600 + 35.36 = 635.36 \, \text{km/hr} \)
- **Westward velocity**: \( 35.36 \, \text{km/hr} \)

### 4. Calculating the Ground Speed:
The resultant ground speed can be found using the Pythagorean theorem, as the plane's movement forms a right triangle:
\[
\text{Ground speed} = \sqrt{(635.36)^2 + (35.36)^2}
\]
\[
= \sqrt{403673.69 + 1249.73} = \sqrt{404923.42} \approx 636.36 \, \text{km/hr}
\]

### 5. Calculating the Angle Off Course:
The angle \( \theta \) off course can be found using the arctangent of the westward component over the northward component:
\[
\theta = \arctan\left(\frac{35.36}{635.36}\right)
\]
\[
= \arctan(0.0556) \approx 3.18^\circ
\]

### Answers:
- **The plane will end up flying approximately 3.18 degrees off course.**
- **The plane's speed relative to the ground will be approximately 636.36 km/hr.**

Would you like further details or have any questions?

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Here are 5 related questions to deepen understanding:
1. How would the ground speed change if the wind speed increased to 100 km/hr?
2. What angle would the plane be off course if the wind came directly from the west?
3. How do you calculate the ground speed when there is a tailwind instead of a crosswind?
4. What effect would a headwind have on the plane’s ground speed?
5. If the plane had to compensate for the wind, what angle should it fly to stay on course?

**Tip:** When analyzing vectors, always resolve them into their components to simplify the calculations for direction and magnitude.

An airplane is heading north at an airspeed of 600 km/hr, but there is a wind blowing from the southwest at 50 km/hr. The plane will end up flying __ degrees off course. The plane's speed relative to the ground will be __ km/hr.

Learn how to calculate the airplane's ground speed and the degree off course when an airplane heads north at 600 km/hr with a 50 km/hr wind blowing from the southwest. This problem involves vector addition, trigonometric functions, and the Pythagorean theorem, suitable for high school math students.

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