Do Now 9/21-9/22... Intro to Proof | Scho... TypingTest.com - C... Westfield High Scho... All Bookmarks Question Show Examples 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 find the height of the skyscraper, we can use trigonometry. Given that Jaxon measures the angle of elevation to the top of the skyscraper from his eye, which is 1.55 meters above the ground, and the horizontal distance from the base of the skyscraper is 266 meters, we can use the tangent function to find the height of the skyscraper. Let's denote: - \( h \) as the height of the skyscraper above the ground. - \( \theta \) as the angle of elevation measured by Jaxon. From the given information: - The angle of elevation \( \theta = 67^{\circ} \). - The horizontal distance from Jaxon to the base of the skyscraper is 266 meters. - The height of Jaxon's eye above the ground is 1.55 meters. We can use the tangent function to find the height of the skyscraper: \[ \tan(\theta) = \frac{h - 1.55}{266} \] Now, we can solve for \( h \) by rearranging the equation: \[ h = 266 \times \tan(67^{\circ}) + 1.55 \] Let's calculate the height of the skyscraper using the tangent function. The height of the skyscraper is approximately 441.066392 meters. Rounding this to the nearest tenth of a meter, the height of the skyscraper is approximately 441.1 meters.