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To solve the problem and find the distance \( BC \), we can break the problem into smaller steps.

1. **Understanding the given dataTo solve the problem and find the distance \( BC \), we can break the problem into smaller steps.

1. **Understanding the given **
   - The angle of elevation of \( A \) from \( C \) is \( 51^\circ \).
   - The angle of depression of \( D \) from \( A \) is \( 28^\circ \).
   - The angle of inclination of the road \( CD \) is \( 34^\circ \).
   - The horizontal distance between \( D \) and the building is \( 33 \) meters.

2. **Determine \( AC \) and \( AD \):**
   - We will use trigonometric relationships to find the lengths of \( AC \) and \( AD \).

Let's start by solving for \( AC \) and \( AD \):

### Step 1: Find \( AC \)

From the right triangle \( \triangle ACD \):
- \( \angle ACD = 34^\circ \)
- \( \angle CAD = 28^\circ \) (since the total angle at \( A \) is \( 90^\circ \))

Using the tangent of \( 34^\circ \):
\[ \tan(34^\circ) = \frac{opposite}{adjacent} = \frac{AD}{33} \]

\[ AD = 33 \times \tan(34^\circ) \]

### Step 2: Find \( AC \)

We have the angle of elevation \( 51^\circ \) and need to find \( AC \):
\[ \tan(51^\circ) = \frac{opposite}{adjacent} = \frac{height}{BC} \]

Given that \( AD \) is vertical and \( BC \) is horizontal, we need to determine \( height \) (or \( AD \)) from the previous step.

### Step 3: Calculate \( BC \)

After determining \( AD \) and the height from \( AC \) using trigonometry:
\[ BC = \sqrt{(AD)^2 + (33)^2} \]

Let's do the calculations:

\[ AD = 33 \times \tan(34^\circ) \]
\[ \tan(34^\circ) \approx 0.6745 \]
\[ AD \approx 33 \times 0.6745 \approx 22.26 \]

For \( AC \):
\[ \tan(51^\circ) = \frac{AC}{BC} \]
\[ \tan(51^\circ) \approx 1.234 \]
\[ BC \approx \frac{AD}{\tan(51^\circ)} \approx \frac{22.26}{1.234} \approx 18.04 \]

So, \( BC \approx 18.04 \) meters. 

### Verification

To verify, let's check if the values match given angles and lengths, and calculate exact \( AD \), and then the hypotenuse \( BC \):

Now, combine the distances for exact \( AC \):

\[ \text{Using the Law of Sines:} \]
\[ \sin(\angle BCA) = \frac{BC}{\sin(\theta_1)} \approx \frac{22.26}{0.813} \approx 27.37 \]

Thus, the distance between \( BC \) is:

\[ \boxed{18.04 \text{ meters}} \]

Would you like any further details on these calculations or other parts of the question?

### Further questions:
1. How can we calculate the length \( CD \) directly?
2. What is the significance of each angle in the problem?
3. How to verify the angle of depression calculation?
4. How can we utilize the ant's speed in the context of part (b)?
5. Can we solve this using an alternate method such as the Law of Cosines?

**Tip:** Always break complex geometric problems into simpler right triangles to apply basic trigonometric functions effectively.

Learn how to solve the problem of finding distance BC in a geometric scenario using trigonometric ratios and right triangle properties. This problem involves angles of elevation and depression, and the Law of Sines. Suitable for high school students.

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