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

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.