 Water main construction. A water main is to be constructed with a 20% grade in the
north direction and a 10% grade in the east direction. Determine the angle θ required in the
water main for the turn from north to east. (Note: The percent grade of a line of slope y/x
is 100(y/x)%. For example, the percent grade of a line of slope 1/2 is 50%.)

To solve this problem, we need to determine the angle \( \theta \) that represents the turn from the north to the east direction, given the percent grades of 20% and 10% respectively.

### Step-by-step solution:

1. **Interpret the percent grades:**
   - A 20% grade in the north direction means the slope is \( \frac{20}{100} = 0.2 \).
   - A 10% grade in the east direction means the slope is \( \frac{10}{100} = 0.1 \).

2. **Understand the geometry of the situation:**
   The two directions (north and east) can be treated as two vectors:
   - The north vector is represented by \( \mathbf{N} = (0, 1, 0.2) \). This is because it's pointing north (along the y-axis) and has a vertical component (z-axis) determined by its slope.
   - The east vector is represented by \( \mathbf{E} = (1, 0, 0.1) \). This is because it's pointing east (along the x-axis) and has a vertical component determined by its slope.

3. **Use the dot product formula to find the angle:**
   The angle \( \theta \) between two vectors is given by the dot product formula:

   \[
   \cos \theta = \frac{\mathbf{N} \cdot \mathbf{E}}{|\mathbf{N}| |\mathbf{E}|}
   \]

   First, calculate the dot product \( \mathbf{N} \cdot \mathbf{E} \):

   \[
   \mathbf{N} \cdot \mathbf{E} = (0 \times 1) + (1 \times 0) + (0.2 \times 0.1) = 0.02
   \]

4. **Find the magnitudes of the vectors:**

   The magnitude of \( \mathbf{N} \) is:

   \[
   |\mathbf{N}| = \sqrt{0^2 + 1^2 + 0.2^2} = \sqrt{1 + 0.04} = \sqrt{1.04} \approx 1.02
   \]

   The magnitude of \( \mathbf{E} \) is:

   \[
   |\mathbf{E}| = \sqrt{1^2 + 0^2 + 0.1^2} = \sqrt{1 + 0.01} = \sqrt{1.01} \approx 1.005
   \]

5. **Calculate \( \cos \theta \):**

   \[
   \cos \theta = \frac{0.02}{1.02 \times 1.005} \approx \frac{0.02}{1.0251} \approx 0.0195
   \]

6. **Find the angle \( \theta \):**

   \[
   \theta = \cos^{-1}(0.0195) \approx 88.88^\circ
   \]

### Conclusion:
The angle \( \theta \) required for the turn from north to east is approximately \( 88.88^\circ \).

---

Would you like any further details or clarifications?

Here are 5 related questions to expand on this concept:
1. How would the angle change if the grade in the east direction were 20% instead of 10%?
2. What if the grade in both directions was 0%, i.e., a flat surface?
3. How do you calculate the vector dot product in 3D space?
4. Can you generalize this method for any two given slopes?
5. How is this problem relevant in designing water pipelines with specific gradients?

**Tip:** The dot product is a powerful tool to calculate the angle between vectors in multidimensional space. Always remember to normalize the vectors before applying the formula.

A water main is to be constructed with a 20% grade in the north direction and a 10% grade in the east direction. Determine the angle θ required in the water main for the turn from north to east.

Determine the angle required for a water main turn from north to east, given a 20% grade in the north direction and a 10% grade in the east direction. This solution explores vector geometry and the dot product formula to calculate the angle, suitable for students in Grades 10-12.

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