9. A surveyor Measures the angle of elevation on the top of a perpendicular building as 19. He moves 120 m nearer the building and finds the angle of elevation is now \( 47^{\circ} \). 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 far is it to the aerial? 11. A flagpole stands on the edge of the top of a building. At a point \( 200 m \) from the buitding the angle ofelevation of the top and bottom of the pole are \( 32^{\circ} \) and \( 30^{\circ} \) respectively. Calculate the height of the flagpole 12. If tan \( x \) a \( =1 \). 5276 , determine sec \( x \), Cosec \( x \) and cot \( x \). CAssume \( x \) is an acute angle

Let's solve each of the problems step-by-step. --- ### **Problem 9** **A surveyor measures the angle of elevation to the top of a perpendicular building as \(19^\circ\). He moves 120 m closer to the building and finds the angle of elevation is now \(47^\circ\). Determine the height of the building.** **Solution:** Let: - \( h \) = height of the building - \( d \) = initial distance from the building From the first position: \[ \tan(19^\circ) = \frac{h}{d} \quad \Rightarrow \quad h = d \cdot \tan(19^\circ) \] After moving 120 m closer: \[ \tan(47^\circ) = \frac{h}{d - 120} \quad \Rightarrow \quad h = (d - 120) \cdot \tan(47^\circ) \] Setting the two equations for \( h \) equal: \[ d \cdot \tan(19^\circ) = (d - 120) \cdot \tan(47^\circ) \] Solving for \( d \): \[ d (\tan(19^\circ) - \tan(47^\circ)) = -120 \cdot \tan(47^\circ) \] \[ d = \frac{120 \cdot \tan(47^\circ)}{\tan(47^\circ) - \tan(19^\circ)} \] Calculating the values: \[ \tan(19^\circ) \approx 0.3443 \] \[ \tan(47^\circ) \approx 1.0724 \] \[ d \approx \frac{120 \times 1.0724}{1.0724 - 0.3443} \approx \frac{128.688}{0.7281} \approx 176.78 \, \text{m} \] Now, finding the height \( h \): \[ h = 176.78 \times 0.3443 \approx 60.8 \, \text{m} \] **Answer:** The height of the building is approximately **61 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?** **Solution:** Let: - \( h = 30 \, \text{m} \) (height of the aerial) - \( d \) = horizontal distance to the aerial Using the tangent function: \[ \tan(32^\circ) = \frac{30}{d} \quad \Rightarrow \quad d = \frac{30}{\tan(32^\circ)} \] Calculating the value: \[ \tan(32^\circ) \approx 0.6249 \] \[ d \approx \frac{30}{0.6249} \approx 48.0 \, \text{m} \] **Answer:** The distance to the aerial is approximately **48 meters**. --- ### **Problem 11** **A flagpole stands on the edge of the top of a building. At a point 200 m from the building, the angles of elevation to the top and bottom of the pole are \(32^\circ\) and \(30^\circ\) respectively. Calculate the height of the flagpole.** **Solution:** Let: - \( H \) = height of the building - \( F \) = height of the flagpole - Distance from the observation point to the building = 200 m From the angles: \[ \tan(30^\circ) = \frac{H}{200} \quad \Rightarrow \quad H = 200 \cdot \tan(30^\circ) \approx 200 \times 0.5774 \approx 115.47 \, \text{m} \] \[ \tan(32^\circ) = \frac{H + F}{200} \quad \Rightarrow \quad H + F = 200 \cdot \tan(32^\circ) \approx 200 \times 0.6249 \approx 124.98 \, \text{m} \] Subtracting the two equations to find \( F \): \[ F = 124.98 \, \text{m} - 115.47 \, \text{m} \approx 9.51 \, \text{m} \] **Answer:** The height of the flagpole is approximately **9.5 meters**. --- ### **Problem 12** **If \( \tan(x) = 1.5276 \), determine \( \sec(x) \), \( \csc(x) \), and \( \cot(x) \). Assume \( x \) is an acute angle.** **Solution:** Given: \[ \tan(x) = 1.5276 \] 1. **Find \( \sec(x) \):** \[ \sec(x) = \frac{1}{\cos(x)} \] First, find \( \cos(x) \): \[ \cos(x) = \frac{1}{\sqrt{1 + \tan^2(x)}} = \frac{1}{\sqrt{1 + (1.5276)^2}} \approx \frac{1}{\sqrt{3.3345}} \approx 0.5472 \] Thus, \[ \sec(x) \approx \frac{1}{0.5472} \approx 1.8305 \] 2. **Find \( \csc(x) \):** \[ \csc(x) = \frac{1}{\sin(x)} = \frac{\sqrt{1 + \tan^2(x)}}{\tan(x)} \approx \frac{1.8255}{1.5276} \approx 1.1943 \] 3. **Find \( \cot(x) \):** \[ \cot(x) = \frac{1}{\tan(x)} = \frac{1}{1.5276} \approx 0.6547 \] **Answers:** - \( \sec(x) \approx 1.8305 \) - \( \csc(x) \approx 1.1973 \) - \( \cot(x) \approx 0.6547 \) ---