The width of a road is b feet. On one side of which there is a window h feet high. A building in front of it subtends an angle `theta` at it, then the height of the building is____

Answer

To find the height of the building (H) given the width of the road (b), the height of the window (h), and the angle subtended by the building at the window (θ), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Geometry**:
- Let the road be represented as line segment AB, where the width of the road is b feet.
- The window is located at point A, and its height is h feet.
- The building is directly in front of the window at point B, subtending an angle θ at point A.

2. **Define the Points**:
- Let point C be the top of the building.
- The height of the building from the ground to point C is H feet.
- The height from the ground to the window (point A) is h feet.

3. **Identify the Angles**:
- The angle subtended at point A by the building is θ.
- The angle at point A from the window to the top of the building can be represented as θ - φ, where φ is the angle of elevation from the window to the top of the building.

4. **Use the Tangent Function**:
- In triangle BDE (where D is the point directly below C on the ground), we can write:
\[ \tan(φ) = \frac{H - h}{b} \]
- This gives us our first equation.

5. **Set Up the Second Triangle**:
- In triangle DEC, we can express:
\[ \tan(θ - φ) = \frac{H - h}{b} \]
- Using the tangent subtraction formula:
\[ \tan(θ - φ) = \frac{\tan(θ) - \tan(φ)}{1 + \tan(θ) \tan(φ)} \]

6. **Substitute the Values**:
- Substitute the expressions for \(\tan(φ)\) and \(\tan(θ)\) into the equation:
\[ \frac{H - h}{b} = \frac{\tan(θ) - \tan(φ)}{1 + \tan(θ) \tan(φ)} \]

7. **Solve for H**:
- Rearranging the equation will yield:
\[ H - h = b \cdot \frac{\tan(θ) - \tan(φ)}{1 + \tan(θ) \tan(φ)} \]
- Finally, express H as:
\[ H = h + b \cdot \frac{\tan(θ) - \tan(φ)}{1 + \tan(θ) \tan(φ)} \]

### Final Formula:
The height of the building can be expressed as:
\[ H = h + \frac{b(\tan(θ) - \tan(φ))}{1 + \tan(θ) \tan(φ)} \]