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com/app/student/3694486/26545770/79af1c669f0ceba6c9bafda2fd1ababo Intro to Proof I Scho.. TypingTest.com-C... Westfield High Scho_ Do Now \( 9 / 21-9 / 22 \) 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. \( \vdots \)

Ask by Mejia Mullins. in the United States
Feb 03,2025

Real Tutor Solution

Tutor-Verified Answer

Answer

The skyscraper is approximately 628.2 meters tall.

Solution

To determine the height of the skyscraper, we can use trigonometric relationships involving right triangles. Here's a step-by-step solution: ### **Given:** - **Height of Jaxon's eye above the ground:** \( 1.55 \) meters - **Horizontal distance from the base of the skyscraper:** \( 266 \) meters - **Angle of elevation to the top of the skyscraper:** \( 67^\circ \) ### **Objective:** Find the total height of the skyscraper. ### **Approach:** 1. **Visual Representation:** Imagine a right triangle where: - The **adjacent side** (base) is the horizontal distance from Jaxon to the skyscraper: \( 266 \) meters. - The **opposite side** is the difference between the skyscraper's height (\( H \)) and Jaxon's eye level (\( 1.55 \) meters). - The **angle of elevation** is \( 67^\circ \). 2. **Trigonometric Relationship:** The tangent of an angle in a right triangle is the ratio of the opposite side to the adjacent side: \[ \tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} \] Plugging in the known values: \[ \tan(67^\circ) = \frac{H - 1.55}{266} \] 3. **Solving for \( H \):** Rearrange the equation to solve for \( H \): \[ H - 1.55 = 266 \times \tan(67^\circ) \] \[ H = 266 \times \tan(67^\circ) + 1.55 \] 4. **Calculating \( \tan(67^\circ) \):** Using a calculator: \[ \tan(67^\circ) \approx 2.35585 \] 5. **Plugging in the Values:** \[ H = 266 \times 2.35585 + 1.55 \] \[ H \approx 626.6551 + 1.55 \] \[ H \approx 628.2051 \text{ meters} \] 6. **Rounding the Answer:** Rounding to the nearest tenth of a meter: \[ H \approx 628.2 \text{ meters} \] ### **Answer:** The height of the skyscraper is approximately **628.2 meters**.

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Beyond the Answer

To find the height of the skyscraper, we can use the tangent function from trigonometry, which relates the angle of elevation to the height and distance. The formula is: \[\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}\] In this case, \(\theta = 67^{\circ}\), the opposite side is the height of the skyscraper minus Jaxon’s eye level (1.55 m), and the adjacent side is the distance from the skyscraper (266 m). Rearranging gives us height = \( \tan(67^{\circ}) \times 266 + 1.55 \) which you can calculate to find the final height. After doing the calculations, the height of the skyscraper would come out to be approximately 325.4 meters when rounded to the nearest tenth!

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