The angle of inclination from the base of skyscraper A to the top of skyscraper \( B \) is approximately \( 128^{\circ} \). If skyscraper B is 1451 feet tall, how far apart are the two skyscrapers? Assume the bases of the two buildings are ai the same elevation. The distance from skyscraper A to skyscraper B is \( \square \) feet. (Round to two decimal places as needed.)

To find the distance between the two skyscrapers, we can use the tangent function from trigonometry. Given: - Angle of inclination from the base of skyscraper A to the top of skyscraper B: \( 128^{\circ} \) - Height of skyscraper B: 1451 feet We can use the tangent function to find the distance between the two skyscrapers. The tangent function is defined as: \[ \tan(\theta) = \frac{\text{opposite side}}{\text{adjacent side}} \] In this case, the opposite side is the height of skyscraper B (1451 feet), and the adjacent side is the distance between the two skyscrapers (which we want to find). Let's denote the distance between the two skyscrapers as \( d \). Using the tangent function, we have: \[ \tan(128^{\circ}) = \frac{1451}{d} \] Now, we can solve for \( d \) by rearranging the equation: \[ d = \frac{1451}{\tan(128^{\circ})} \] Let's calculate the value of \( d \) using the tangent function. The distance between the two skyscrapers is approximately 1394.37 feet. Therefore, the distance between the two skyscrapers is 1394.37 feet.