A far off planet is estimated to be at a distance D from the earth. If its diametrically opposite extremes subtend an angle `theta` at an observatory situated on the earth, the approximate diameter of the planet is

Options

A. `theta/D`
B. `D/theta`
C. `D theta`
D. `1/(D theta)`

Answer

To solve the problem of finding the approximate diameter of a far-off planet that subtends an angle \( \theta \) at an observatory on Earth, we can follow these steps:

### Step 1: Understand the Geometry
We have an observatory on Earth and a planet at a distance \( D \). The planet's diametrically opposite extremes subtend an angle \( \theta \) at the observatory. We can visualize this scenario as a triangle formed by the observatory and the two extremes of the planet.

### Step 2: Define the Variables
Let:
- \( D \) = distance from the observatory to the planet
- \( d \) = diameter of the planet
- The angle subtended at the observatory is \( \theta \).

### Step 3: Consider Half of the Angle
Since the diameter subtends the angle \( \theta \), each half of the diameter subtends an angle of \( \frac{\theta}{2} \).

### Step 4: Use Trigonometric Relationships
In the triangle formed (let's label the observatory point as \( O \), one extreme of the planet as \( A \), and the other extreme as \( C \)), we can use the tangent function:
\[ \tan\left(\frac{\theta}{2}\right) = \frac{\text{opposite}}{\text{adjacent}} = \frac{\frac{d}{2}}{D} \]
Where \( \frac{d}{2} \) is the opposite side (half the diameter of the planet) and \( D \) is the adjacent side (distance from the observatory to the planet).

### Step 5: Approximate for Small Angles
For small angles (which is the case here), we can use the approximation:
\[ \tan\left(\frac{\theta}{2}\right) \approx \frac{\theta}{2} \]
Thus, we can rewrite the equation as:
\[ \frac{\theta}{2} = \frac{\frac{d}{2}}{D} \]

### Step 6: Solve for the Diameter \( d \)
Rearranging the equation gives:
\[ \frac{d}{2} = D \cdot \frac{\theta}{2} \]
Multiplying both sides by 2:
\[ d = D \cdot \theta \]

### Conclusion
The approximate diameter of the planet is:
\[ d \approx D \cdot \theta \]

### Final Answer
Thus, the approximate diameter of the planet is \( d = D \cdot \theta \).

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