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### Problem 16:
A cliff 1 km long is represented on a map of scale 2 cm to 1 km by a line of length 0.42 cm. What is the average inclination of the cliff to the horizontal?

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**Solution:**

1. **Convert the map scale to real-world length:**
   - The map's scale is \( 2 \, \text{cm} = 1 \, \text{km} \).
   - The line on the map is \( 0.42 \, \text{cm} \).
   - The real-world length corresponding to this line is:
     \[
     \frac{0.42 \, \text{cm}}{2 \, \text{cm}} = 0.21 \, \text{km}
     \]

2. **Calculate the average inclination:**
   The real-world horizontal distance (length of the cliff) is 1 km, and the vertical rise (from the map) is 0.21 km.

   Using the formula for the inclination angle \( \theta \):
   \[
   \tan(\theta) = \frac{\text{vertical rise}}{\text{horizontal distance}} = \frac{0.21}{1}
   \]
   \[
   \theta = \tan^{-1}(0.21) \approx 11.86^\circ
   \]

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### Problem 17:
A man walks 100 m up a slope of 14° and then 50 m up a slope of 12°. How much higher is he than when he started?

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**Solution:**

1. **Height gained on the first slope:**
   The height gained can be calculated using the formula \( h = d \sin(\theta) \), where \( d \) is the distance walked and \( \theta \) is the angle of inclination.
   
   - First slope: \( d_1 = 100 \, \text{m} \), \( \theta_1 = 14^\circ \)
     \[
     h_1 = 100 \sin(14^\circ) \approx 100 \times 0.2419 = 24.19 \, \text{m}
     \]

2. **Height gained on the second slope:**
   - Second slope: \( d_2 = 50 \, \text{m} \), \( \theta_2 = 12^\circ \)
     \[
     h_2 = 50 \sin(12^\circ) \approx 50 \times 0.2079 = 10.39 \, \text{m}
     \]

3. **Total height gained:**
   \[
   \text{Total height} = h_1 + h_2 = 24.19 \, \text{m} + 10.39 \, \text{m} = 34.58 \, \text{m}
   \]

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### Problem 18:
The bob of a pendulum 4 m long is 25 cm higher at the top of its swing than it is at the bottom. Find the angle of swing on each side of the vertical.

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**Solution:**

1. **Pendulum height change:**
   The change in height \( h = 25 \, \text{cm} = 0.25 \, \text{m} \).

2. **Relationship between height and angle:**
   The height change \( h \) is related to the length of the pendulum \( L \) and the angle \( \theta \) by:
   \[
   h = L (1 - \cos(\theta))
   \]
   Solving for \( \theta \):
   \[
   0.25 = 4(1 - \cos(\theta))
   \]
   \[
   1 - \cos(\theta) = \frac{0.25}{4} = 0.0625
   \]
   \[
   \cos(\theta) = 1 - 0.0625 = 0.9375
   \]
   \[
   \theta = \cos^{-1}(0.9375) \approx 20.87^\circ
   \]

Thus, the angle of swing on each side of the vertical is approximately \( 20.87^\circ \).

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Would you like further details or explanations on any of these problems? Here are some additional related questions you might explore:

1. How would you calculate the real-world length if the map scale were different?
2. What is the maximum height the man can gain if he keeps walking up slopes of increasing angles?
3. How does the period of the pendulum change with the length or amplitude of the swing?
4. How can you calculate the horizontal distance traveled by the man in Problem 17?
5. How does the tangent function help in determining inclinations in real-world applications?

**Tip:** When working with angles and slopes, always keep trigonometric ratios in mind, as they provide easy access to height and distance relations.

16. A cliff 1 km long is represented on a map of scale 2 cm to 1 km by a line of length 0.42 cm. What is the average inclination of the cliff to the horizontal? 
17. A man walks 100 m up a slope of 14° and then 50 m up a slope of 12°. How much higher is he than when he started? 
18. The bob of a pendulum 4 m long is 25 cm higher at the top of its swing than it is at the bottom. Find the angle of swing on each side of the vertical.

This page covers the solutions to three trigonometry-related problems: determining the average inclination of a cliff from a map, calculating the height gained while walking up slopes, and finding the angle of swing of a pendulum. Suitable for high school students in Grades 10-12, this guide walks through the use of tangent, sine, and cosine functions.

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