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Let's break down the problem:

### Given:
1. The plane is flying at a velocity of 75.0 km/hr [N 65.0° E].
2. The wind is blowing from the northwest at a speed of 45.0 km/hr, in a direction 55.0° below the horizontal.

### Part (a): Sketch the Scalene Triangle

You need to sketch a triangle that represents the situation using a scale of **1 cm = 5 km/hr**. The triangle will have three sides:
- **Side P** representing the plane's velocity (75.0 km/hr)
- **Side W** representing the wind's velocity (45.0 km/hr)
- **Side G** representing the ground velocity (unknown at this point)

### Steps to Sketch:
1. **Draw Side P (Plane's Velocity):** 
   - Measure out 75.0 km/hr using the scale, which is \( 75.0 \, \text{km/hr} \times \frac{1 \, \text{cm}}{5 \, \text{km/hr}} = 15 \, \text{cm} \).
   - Draw this line at an angle of 65.0° East of North.

2. **Draw Side W (Wind's Velocity):** 
   - Measure out 45.0 km/hr using the scale, which is \( 45.0 \, \text{km/hr} \times \frac{1 \, \text{cm}}{5 \, \text{km/hr}} = 9 \, \text{cm} \).
   - Since the wind is blowing from the northwest at 55° below the horizontal, you should draw this line at an angle of 125° from the east (since northwest is 45° from the north and 55° below that is 125°).

3. **Label the triangle:**
   - The vertex where the plane's velocity vector starts should be labeled **P**.
   - The end of the wind vector should be **W**.
   - The ground vector should be labeled **G**.

### Part (b): Cosine/Sine Law Method

Now, you need to determine the velocity of the plane relative to the ground using the **cosine law** or **sine law**.

#### Cosine Law:
For any triangle:
\[
c^2 = a^2 + b^2 - 2ab\cos(C)
\]
Where \( c \) is the side opposite the angle \( C \).

In this scenario:
- \( a = 75.0 \, \text{km/hr} \) (Plane's velocity)
- \( b = 45.0 \, \text{km/hr} \) (Wind's velocity)
- \( C = 180^\circ - (65^\circ + 55^\circ) = 60^\circ \)

Now apply the cosine law:
\[
G^2 = 75.0^2 + 45.0^2 - 2 \times 75.0 \times 45.0 \times \cos(60^\circ)
\]
\[
G^2 = 5625 + 2025 - 2 \times 75 \times 45 \times 0.5
\]
\[
G^2 = 7650 - 3375
\]
\[
G = \sqrt{4275} \, \text{km/hr}
\]
\[
G \approx 65.38 \, \text{km/hr}
\]

Now, determine the angle using the sine law if needed.

Would you like further details on the process or help with the sine law to find the exact direction? Here are some related questions to explore this concept further:

1. How does the angle of the wind relative to the plane affect the ground velocity?
2. What would happen if the wind were blowing directly opposite to the plane's direction?
3. How can the sine law be used to find the angle of the ground velocity?
4. Why is the cosine law preferred in this scenario?
5. How does the resultant ground velocity change if the plane's velocity increases?

**Tip:** When dealing with vectors, always ensure your angles are measured correctly, especially when working with non-right triangles like in this scenario.

The pilot of a small plane checks her readings and sees she is flying at a velocity of 75.0 km/hr [N 65.0° E] according to the heavy wind around the plane. According to a ground observer, the heavy wind is blowing from the north-west (pointing along the direction of 55.0° below the horizontal) at a speed of 45.0 km/hr. Using the cosine law/sine law method, determine the velocity of the plane (in km/hr) relative to the ground, indicating both magnitude and angle from the horizontal axis.

Learn how to determine the ground velocity of a plane flying at 75.0 km/hr [N 65.0° E] while facing a strong wind from the northwest at 45.0 km/hr. This problem involves using the Cosine and Sine Law for vector addition to find the magnitude and direction of the resultant velocity.

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