A man 2 metres tall walks towards a mast of height 26metres at the rate of 1.67m/s determine the rate of change  of the angle of elevation of the top of man's head when he is 16 metres from the mast

To solve this problem, we will use related rates and trigonometry.

Let:
- \( h = 26 \) meters be the height of the mast.
- \( y = 2 \) meters be the height of the man.
- \( x(t) \) be the distance between the man and the base of the mast at time \( t \).
- \( \theta(t) \) be the angle of elevation from the top of the man's head to the top of the mast.

### Step-by-Step Solution

1. **Define the right triangle:**
   The distance from the man to the base of the mast is \( x(t) \), and the height difference between the top of the mast and the man's head is \( h - y = 26 - 2 = 24 \) meters.

   The angle of elevation \( \theta \) is given by:
   \[
   \tan(\theta) = \frac{24}{x(t)}.
   \]

2. **Differentiate with respect to time \( t \):**

   To find the rate of change of the angle \( \frac{d\theta}{dt} \), we take the derivative of both sides with respect to \( t \):
   \[
   \frac{d}{dt} \left( \tan(\theta) \right) = \frac{d}{dt} \left( \frac{24}{x(t)} \right).
   \]

   Using the chain rule on the left-hand side:
   \[
   \sec^2(\theta) \cdot \frac{d\theta}{dt} = -\frac{24}{x(t)^2} \cdot \frac{dx}{dt}.
   \]

3. **Substitute given values:**

   At the moment when \( x = 16 \) meters, and the rate of change of distance is \( \frac{dx}{dt} = -1.67 \) m/s (negative because the distance is decreasing).

   Calculate \( \tan(\theta) \) when \( x = 16 \):
   \[
   \tan(\theta) = \frac{24}{16} = \frac{3}{2}.
   \]

   Therefore:
   \[
   \theta = \tan^{-1}\left( \frac{3}{2} \right).
   \]

   Also, we have:
   \[
   \sec^2(\theta) = 1 + \tan^2(\theta) = 1 + \left( \frac{3}{2} \right)^2 = 1 + \frac{9}{4} = \frac{13}{4}.
   \]

4. **Solve for \( \frac{d\theta}{dt} \):**

   Substitute these values into the differentiated equation:
   \[
   \frac{13}{4} \cdot \frac{d\theta}{dt} = -\frac{24}{16^2} \cdot (-1.67).
   \]

   Simplify:
   \[
   \frac{13}{4} \cdot \frac{d\theta}{dt} = \frac{24 \times 1.67}{256}.
   \]

   Solve for \( \frac{d\theta}{dt} \):
   \[
   \frac{d\theta}{dt} = \frac{24 \times 1.67}{256} \cdot \frac{4}{13}.
   \]

   Now, calculate the numeric value:
   \[
   \frac{d\theta}{dt} \approx \frac{40.08}{256} \cdot \frac{4}{13}.
   \]

   Simplify:
   \[
   \frac{d\theta}{dt} \approx \frac{40.08 \times 4}{256 \times 13} = \frac{160.32}{3328} \approx 0.0482 \, \text{radians per second}.
   \]

### Final Answer:

The rate of change of the angle of elevation of the top of the man's head is approximately \( 0.0482 \) radians per second when he is 16 meters from the mast.

Would you like more details or have any questions?

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Here are 5 related questions for further exploration:

1. How would the rate change if the height of the mast were different?
2. What would be the rate of change if the man were walking away from the mast?
3. How does the height of the man affect the rate of change of the angle?
4. What is the significance of the rate of change being positive or negative in this context?
5. How would you apply related rates to a situation involving a moving car and a stationary observer?

**Tip:** Understanding the relationship between angles and distances can help solve various real-life problems, from navigation to architectural design.

A man 2 metres tall walks towards a mast of height 26 metres at the rate of 1.67m/s. Determine the rate of change of the angle of elevation of the top of the man's head when he is 16 metres from the mast.

Explore how to calculate the rate of change of the angle of elevation for a man walking towards a 26m tall mast. Using related rates, trigonometry, and differentiation, this problem demonstrates how the angle changes when the man is 16 meters from the mast, walking at 1.67 m/s. Suitable for Grades 11-12 students.

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