What is the angle subtended by 1 m pole at distance 1 km on the ground in sexagesimal measure?

Options

A. A) `(9)/( 50 pi) ` deg ree
B. B) `( 9)/(5pi ) ` deg ree
C. C) 3.4 min
D. D) 3.5 min

Answer

To find the angle subtended by a 1 m pole at a distance of 1 km on the ground in sexagesimal measure, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Problem**: We need to find the angle θ subtended by a 1 m pole at a distance of 1 km from the base of the pole.

2. **Convert Distance**: Convert the distance from kilometers to meters.
\[ 1 \text{ km} = 1000 \text{ m} \]

3. **Identify the Right Triangle**: In this scenario, we can visualize a right triangle where:
- The height of the pole (perpendicular) = 1 m
- The distance from the base of the pole to the point on the ground (base) = 1000 m

4. **Use the Tangent Function**: The tangent of the angle θ can be defined as the ratio of the opposite side (height of the pole) to the adjacent side (distance from the pole).
\[ \tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{1}{1000} \]

5. **Calculate the Angle**: To find θ, we take the inverse tangent (arctan) of the ratio.
\[ \theta = \tan^{-1}\left(\frac{1}{1000}\right) \]

6. **Approximate for Small Angles**: For very small angles, we can use the approximation that \( \tan \theta \approx \theta \) (in radians). Thus,
\[ \theta \approx \frac{1}{1000} \text{ radians} \]

7. **Convert Radians to Degrees**: To convert from radians to degrees, we use the conversion factor \( \frac{180}{\pi} \).
\[ \theta \text{ (in degrees)} = \theta \text{ (in radians)} \times \frac{180}{\pi} \]
Substituting the value:
\[ \theta \approx \frac{1}{1000} \times \frac{180}{\pi} \]

8. **Calculate the Final Value**:
\[ \theta \approx \frac{180}{1000 \pi} \approx \frac{180}{3141.59} \approx 0.0573 \text{ degrees} \]

9. **Simplify the Result**:
\[ \theta \approx \frac{9}{50} \text{ degrees} \text{ (after simplification)} \]

### Final Answer:
The angle subtended by the 1 m pole at a distance of 1 km on the ground is approximately \( \frac{9}{50} \) degrees.