A student measures the angle of elevation of burw from two different points from point $$ P $$ and $$ D_{1} $$ po peat the angle of elevation of the top of the bunsectively of the distance between $$ P $$ lo $$ \& 3 $$ 1545 m , calculate the beight of the bulding

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It appears there might be some typographical errors in your problem statement. However, based on the context, I'll interpret the problem as follows:

**Problem Interpretation:**
- A student measures the angles of elevation to the top of a building from two different points, $$ P $$ and $$ D_1 $$.
- The distance between points $$ P $$ and $$ D_1 $$ is $$ 1545 $$ meters.
- Let’s assume the angles of elevation from $$ P $$ and $$ D_1 $$ are $$ \theta $$ and $$ \phi $$ degrees, respectively.
- The goal is to calculate the height $$ h $$ of the building.

**Solution:**

To find the height of the building, we'll use trigonometric principles involving right triangles formed by the observation points, the base of the building, and the top of the building.

1. **Define the Variables:**
   - Let $$ h $$ be the height of the building.
   - Let $$ x $$ be the horizontal distance from point $$ P $$ to the base of the building.
   - Let $$ y $$ be the horizontal distance from point $$ D_1 $$ to the base of the building.
   - Given that the distance between $$ P $$ and $$ D_1 $$ is $$ 1545 $$ meters, we have:
     $$
     x + y = 1545 \text{ meters}
     $$

2. **Apply the Tangent Function:**
   - From point $$ P $$:
     $$
     \tan(\theta) = \frac{h}{x} \quad \Rightarrow \quad x = \frac{h}{\tan(\theta)}
     $$
   - From point $$ D_1 $$:
     $$
     \tan(\phi) = \frac{h}{y} \quad \Rightarrow \quad y = \frac{h}{\tan(\phi)}
     $$

3. **Set Up the Equation:**
   - Since $$ x + y = 1545 $$ meters:
     $$
     \frac{h}{\tan(\theta)} + \frac{h}{\tan(\phi)} = 1545
     $$
   - Factor out $$ h $$:
     $$
     h \left( \frac{1}{\tan(\theta)} + \frac{1}{\tan(\phi)} \right) = 1545
     $$
   - Simplify using the identity $$ \frac{1}{\tan(\alpha)} = \cot(\alpha) $$:
     $$
     h (\cot(\theta) + \cot(\phi)) = 1545
     $$
   - Solve for $$ h $$:
     $$
     h = \frac{1545}{\cot(\theta) + \cot(\phi)}
     $$
   - Alternatively, using the fact that $$ \cot(\alpha) = \frac{\tan(\beta)}{\tan(\beta)\tan(\alpha)} $$, you can express $$ h $$ as:
     $$
     h = \frac{1545 \times \tan(\theta) \times \tan(\phi)}{\tan(\theta) + \tan(\phi)}
     $$

**Final Formula:**
$$
h = \frac{1545 \cdot \tan(\theta) \cdot \tan(\phi)}{\tan(\theta) + \tan(\phi)}
$$

**Example Calculation:**

Suppose the angles of elevation are:
- $$ \theta = 30^\circ $$
- $$ \phi = 45^\circ $$

1. Calculate the tangents:
   $$
   \tan(30^\circ) = \frac{\sqrt{3}}{3} \approx 0.5774
   $$
   $$
   \tan(45^\circ) = 1
   $$
2. Plug into the formula:
   $$
   h = \frac{1545 \times 0.5774 \times 1}{0.5774 + 1} = \frac{1545 \times 0.5774}{1.5774} \approx \frac{892.8}{1.5774} \approx 566 \text{ meters}
   $$

**Conclusion:**

To determine the height of the building, you need the specific angles of elevation from both points $$ P $$ and $$ D_1 $$. Once you have those angles, you can apply the formula above to calculate the height.

If you can provide the exact angles of elevation, I can help you compute the numerical value of the building's height.
To find the height of the building, use the formula: $$ h = \frac{1545 \times \tan(\theta) \times \tan(\phi)}{\tan(\theta) + \tan(\phi)} $$ Where $$ \theta $$ and $$ \phi $$ are the angles of elevation from points $$ P $$ and $$ D_1 $$ respectively.

A student measures the angle of elevation of burw from two different points from point \( P \) and \( D_{1} \) po peat the angle of elevation of the top of the bunsectively of the distance between \( P \) lo \( \& 3 \) 1545 m , calculate the beight of the bulding

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