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The problem in the image involves a plane flying at a constant horizontal altitude of 3800 m. It asks for the rate of change of distance \( s \) between the plane and a fixed point \( P \) on the ground as the angle \( \theta \) formed between the plane's line of sight and the horizontal line from point \( P \) is \( 30^\circ \).

### Solution Outline:

This is a related rates problem involving trigonometry. Given that the plane is flying horizontally at an altitude of 3800 m and the angle \( \theta \) changes with respect to time, we need to find how the distance \( s \) (hypotenuse) changes with time when \( \theta = 30^\circ \).

#### Step-by-Step Approach:

1. **Relation between variables**: 
   From the diagram, we see a right triangle where:
   - The vertical leg is the constant height of 3800 m.
   - The hypotenuse is \( s \), the distance from the plane to point \( P \).
   - The angle \( \theta \) is the angle between the horizontal ground and the hypotenuse.

   We can use the relation:
   \[
   \sin(\theta) = \frac{3800}{s}
   \]
   Therefore, the distance \( s \) is:
   \[
   s = \frac{3800}{\sin(\theta)}
   \]

2. **Differentiating with respect to time**:
   To find the rate of change of \( s \) with respect to time (\( \frac{ds}{dt} \)), we differentiate the equation implicitly with respect to \( t \):
   \[
   \frac{ds}{dt} = \frac{-3800 \cdot \cos(\theta)}{\sin^2(\theta)} \cdot \frac{d\theta}{dt}
   \]

3. **Substituting known values**:
   We are given that \( \theta = 30^\circ \) and need to compute \( \frac{ds}{dt} \) at this angle. Plugging in the trigonometric values:
   \[
   \sin(30^\circ) = \frac{1}{2}, \quad \cos(30^\circ) = \frac{\sqrt{3}}{2}
   \]

   Substituting these into the equation:
   \[
   \frac{ds}{dt} = \frac{-3800 \cdot \frac{\sqrt{3}}{2}}{\left(\frac{1}{2}\right)^2} \cdot \frac{d\theta}{dt}
   \]

   Simplifying:
   \[
   \frac{ds}{dt} = -3800 \cdot \frac{\sqrt{3}}{2} \cdot 4 \cdot \frac{d\theta}{dt} = -3800 \cdot 2\sqrt{3} \cdot \frac{d\theta}{dt}
   \]
   Therefore, the rate of change of \( s \) is:
   \[
   \frac{ds}{dt} = -7600\sqrt{3} \cdot \frac{d\theta}{dt}
   \]

To find the numerical value of \( \frac{ds}{dt} \), we would need the value of \( \frac{d\theta}{dt} \), which represents the rate of change of the angle \( \theta \) with respect to time.

Would you like more details on any part, or help calculating further based on additional information?

### Follow-up Questions:
1. What is the value of \( \frac{d\theta}{dt} \) given in the problem?
2. Would you like an explanation of implicit differentiation used in this solution?
3. How would you calculate the rate of change if the plane's horizontal speed was given?
4. Do you want to understand the trigonometric identities used in this context?
5. Can we assume the rate of change of angle is constant, or does it vary over time?

### Tip:
Related rates problems often involve implicit differentiation, where every variable is treated as a function of time. Remember to differentiate every term with respect to time and account for the chain rule.

Pesawat terbang pada lintasan horizontal pada ketinggian 3800 m. Berapakah laju perubahan jarak s antara pesawat tersebut dengan titik tetap P terhadap teta pada saat teta = 30 derajat?

Learn how to solve a related rates problem involving a plane flying at an altitude of 3800 m. This solution includes step-by-step differentiation using trigonometric functions to find the rate of change of distance as the angle changes. Suitable for high school students in Grades 10-12.

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