9. A surveyor Measures the angle of eleration on the top of a perpendicular building as \( 19^{\circ} \). He moves 120 M nearer the buitding and finds the angle of elevation is now 47 Determine the height of the building. 10. If the angle of elevation of the top of a vertical 30 m high aerial is \( 32^{\circ} \), how for is it to the aerial? 1. A flagpole stands on the edge of the top a building. At a point 200 m from the build the angle of elevation of the top and both of the pole are \( 32^{\circ} \) and \( 30^{\circ} \) respectively. Calculate the height of the flagpole.

Alright, I need to solve these three trigonometry problems involving angles of elevation and heights of buildings or flagpoles. Let's tackle them one by one. **Problem 9:** A surveyor measures the angle of elevation on the top of a perpendicular building as \(19^{\circ}\). He moves 120 meters nearer the building and finds the angle of elevation is now \(47^{\circ}\). Determine the height of the building. First, let's visualize the scenario. There's a building with a certain height, and the surveyor is initially at a distance from the building. He measures the angle of elevation to the top of the building as \(19^{\circ}\). Then, he moves 120 meters closer to the building and measures the angle of elevation as \(47^{\circ}\). I need to find the height of the building. Let's denote: - \( h \) as the height of the building. - \( d \) as the initial distance from the surveyor to the building. - \( d - 120 \) as the new distance after moving 120 meters closer. Using the tangent function, which relates the angle of elevation to the opposite side (height) and the adjacent side (distance), we can set up two equations based on the two measurements. First measurement: \[ \tan(19^{\circ}) = \frac{h}{d} \] So, \[ h = d \cdot \tan(19^{\circ}) \] Second measurement: \[ \tan(47^{\circ}) = \frac{h}{d - 120} \] So, \[ h = (d - 120) \cdot \tan(47^{\circ}) \] Since both expressions equal \( h \), we can set them equal to each other: \[ d \cdot \tan(19^{\circ}) = (d - 120) \cdot \tan(47^{\circ}) \] Now, let's solve for \( d \): \[ d \cdot \tan(19^{\circ}) = d \cdot \tan(47^{\circ}) - 120 \cdot \tan(47^{\circ}) \] \[ d \cdot \tan(19^{\circ}) - d \cdot \tan(47^{\circ}) = -120 \cdot \tan(47^{\circ}) \] \[ d (\tan(19^{\circ}) - \tan(47^{\circ})) = -120 \cdot \tan(47^{\circ}) \] \[ d = \frac{-120 \cdot \tan(47^{\circ})}{\tan(19^{\circ}) - \tan(47^{\circ})} \] Calculating the values: \[ \tan(19^{\circ}) \approx 0.3443 \] \[ \tan(47^{\circ}) \approx 1.0724 \] \[ d = \frac{-120 \cdot 1.0724}{0.3443 - 1.0724} = \frac{-128.688}{-0.7281} \approx 176.5 \text{ meters} \] Now, plug \( d \) back into the first equation to find \( h \): \[ h = 176.5 \cdot 0.3443 \approx 60.8 \text{ meters} \] So, the height of the building is approximately 60.8 meters. **Problem 10:** If the angle of elevation of the top of a vertical 30 m high aerial is \(32^{\circ}\), how far is it to the aerial? This problem is straightforward. We have the height of the aerial and the angle of elevation, and we need to find the distance to the aerial. Using the tangent function: \[ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{h}{d} \] Where: - \( h = 30 \) meters (height of the aerial) - \( \theta = 32^{\circ} \) (angle of elevation) - \( d \) is the distance to the aerial. Rearranging the formula to solve for \( d \): \[ d = \frac{h}{\tan(\theta)} = \frac{30}{\tan(32^{\circ})} \] Calculating \( \tan(32^{\circ}) \): \[ \tan(32^{\circ}) \approx 0.6249 \] \[ d = \frac{30}{0.6249} \approx 48 \text{ meters} \] So, the aerial is approximately 48 meters away. **Problem 1.** A flagpole stands on the edge of the top of a building. At a point 200 m from the building, the angle of elevation of the top and both of the pole are \(32^{\circ}\) and \(30^{\circ}\) respectively. Calculate the height of the flagpole. This problem involves two angles of elevation: one for the top of the building and another for the top of the flagpole. Let's denote: - \( h_b \) as the height of the building. - \( h_f \) as the height of the flagpole. - \( d = 200 \) meters as the distance from the point to the building. First, consider the angle of elevation to the top of the building: \[ \tan(32^{\circ}) = \frac{h_b}{200} \] \[ h_b = 200 \cdot \tan(32^{\circ}) \approx 200 \cdot 0.6249 = 124.98 \text{ meters} \] Next, consider the angle of elevation to the top of the flagpole: \[ \tan(30^{\circ}) = \frac{h_b + h_f}{200} \] \[ h_b + h_f = 200 \cdot \tan(30^{\circ}) \approx 200 \cdot 0.5774 = 115.48 \text{ meters} \] But wait, this doesn't make sense because \( h_b \) is already calculated as approximately 124.98 meters, which is greater than 115.48 meters. This indicates a contradiction, suggesting that there might be an error in the problem statement or the given angles. Perhaps the angles are meant to be for different points or there's a misinterpretation. Alternatively, if the flagpole is attached to the building, the angles should align in a way that the sum of heights doesn't exceed the calculated building height. Given the current information, it's impossible to determine the height of the flagpole without additional data or clarification.