To solve this problem, we need to determine the slope distance when the horizontal distance is 325.75 meters and the slope is given as 13 minutes and 6 seconds (denoted as $13' 06''$).

Formula:

The relationship between the horizontal distance $D_h$, slope distance $D_s$, and the vertical angle $\theta$ is:

$D_s = \frac{D_h}{\cos(\theta)}$

Where:

• $D_s$ = Slope distance (what we are solving for)
• $D_h$ = Horizontal distance (325.75 meters)
• $\theta$ = Slope angle

Step 1: Convert angle to degrees

The given angle is $13' 06''$. We convert this into decimal degrees: $\theta = 13 + \frac{6}{60} = 13 + 0.1 = 13.1 \text{ minutes}$ Now, convert minutes into degrees: $\theta = \frac{13.1}{60} \text{ degrees} \approx 0.21833 \text{ degrees}$

Step 2: Calculate the cosine of the angle

$\cos(\theta) = \cos(0.21833^\circ)$

Step 3: Apply the formula

Finally, plug the values into the formula to get the slope distance:

$D_s = \frac{325.75}{\cos(0.21833^\circ)}$

I'll calculate this next.The slope distance that should be laid out is approximately 325.75 meters.

It appears that the small angle had a negligible effect on the distance, resulting in a slope distance almost identical to the horizontal distance.

Would you like further details or have any questions? Here are 5 related questions to consider:

1. How is the vertical angle typically measured in surveying?
2. What are other ways to calculate slope distance if the angle is larger?
3. How would the result change if the angle were in degrees instead of minutes and seconds?
4. Why is it important to account for slope distance in practical surveying applications?
5. What is the difference between horizontal and slope distance in field measurements?

Tip: Small angles like the one in this problem often lead to negligible differences between horizontal and slope distances, but for larger angles, the difference becomes more significant.