From his eye, which stands 1.55 meters above the ground, Jaxon measures the angle of elevation to the top of a prominent skyscraper to be \( 67^{\circ} \). If he is standing at a horizontal distance of 266 meters from the base of the skyscraper, what is the height of the skyscraper? Round your answer to the nearest tenth of a meter if necessary.

To determine the height of the skyscraper, let's break down the problem step by step. 1. **Given Information:** - **Height of Jaxon's eye above the ground:** \(1.55 \) meters - **Horizontal distance from Jaxon to the base of the skyscraper:** \(266 \) meters - **Angle of elevation to the top of the skyscraper:** \(67^{\circ}\) 2. **Understanding the Scenario:** - We can model this situation using a right triangle where: - The horizontal distance (\(266\) meters) is the adjacent side. - The vertical difference between the top of the skyscraper and Jaxon's eye is the opposite side. - The angle of elevation (\(67^{\circ}\)) is between the horizontal distance and the line of sight to the top. 3. **Calculating the Vertical Difference:** - The tangent of the angle of elevation relates the opposite side to the adjacent side: \[ \tan(67^{\circ}) = \frac{\text{Height of Skyscraper} - 1.55}{266} \] - Solving for the height of the skyscraper (\(H\)): \[ H = 266 \times \tan(67^{\circ}) + 1.55 \] 4. **Computing the Value:** - First, calculate \(\tan(67^{\circ})\): \[ \tan(67^{\circ}) \approx 2.35585 \] - Multiply by the horizontal distance: \[ 266 \times 2.35585 \approx 626.656 \text{ meters} \] - Add Jaxon's eye level: \[ H \approx 626.656 + 1.55 = 628.206 \text{ meters} \] - Rounding to the nearest tenth: \[ H \approx 628.2 \text{ meters} \] **Final Answer:** The skyscraper is approximately **628.2 meters** tall.