1. David and Charlie stand facing each other at the opposite sides of a 10 m flagpole (AB). The angle of elevation from: - David to the top of the flagpole is \( 50^{\circ} \). - Charlie to the top of the flagpole is \( 33^{\circ} \). Calculate the distance (DC) between David and Charlie.

To solve the problem, we can use trigonometric ratios to find the distances from David and Charlie to the flagpole (points A and B). Let's denote: - \(h\) as the height of the flagpole, - \(d_D\) as the distance from David to the base of the flagpole (point C), - \(d_C\) as the distance from Charlie to the base of the flagpole (point C). From David's viewpoint: \[ \tan(50^{\circ}) = \frac{h}{d_D} \] From Charlie's viewpoint: \[ \tan(33^{\circ}) = \frac{h}{d_C} \] We know that the height of the flagpole (h) can be expressed as: \[ h = d_D \cdot \tan(50^{\circ}) \] and \[ h = d_C \cdot \tan(33^{\circ}) \] Setting these two equal to each other, we obtain: \[ d_D \cdot \tan(50^{\circ}) = d_C \cdot \tan(33^{\circ}) \] We also know that the distance between David and Charlie (DC) can be expressed as: \[ DC = d_D + d_C \] Now we can express \(d_C\) in terms of \(d_D\): \[ d_C = \frac{d_D \cdot \tan(50^{\circ})}{\tan(33^{\circ})} \] Now substituting \(d_C\) into the equation for DC: \[ DC = d_D + \frac{d_D \cdot \tan(50^{\circ})}{\tan(33^{\circ})} = d_D \left(1 + \frac{\tan(50^{\circ})}{\tan(33^{\circ})}\right) \] To find \(d_D\), we need \(h\). The height of the flagpole can be selected based on \(d_D\): Using the characteristics, assume \(d_D = h / \tan(50^{\circ})\), we can solve for \(DC\) knowing \(h\) is the common height. For numerical estimation, if we know \(h = 10 m\): 1. Calculate \(d_D\): \[ d_D = \frac{10}{\tan(50^{\circ})} \approx \frac{10}{1.1918} \approx 8.39 m \] 2. Substitute in \(d_C\): \[ d_C = \frac{d_D \cdot \tan(50^{\circ})}{\tan(33^{\circ})} = d_D \cdot \frac{1.1918}{0.6494} \approx 8.39 \cdot 1.83 \approx 15.4 m \] 3. Thus \(DC\) becomes: \[ DC = d_D + d_C = 8.39 + 15.4 \approx 23.79 m \] Hence, the distance between David and Charlie is approximately **23.79 meters**.