Question

A tourist standing in the center of the roof of a 1,000-foot skyscraper wants to take a photograph of the top of an adjacent skyscraper that is 1,550 feet tall.

If the centers of the skyscrapers are 1,050 feet apart, approximate the angle of elevation from the tourist to the top of the adjacent skyscraper to the nearest tenth of a degree.
55.9°
27.6°
57.2°
34.1°

Answer 1

9514 1404 393

Answer:

27.6°

Explanation:

The horizontal distance to the adjacent skyscraper is 1050 feet, and the vertical distance to the top is 550 feet. These distances are the adjacent and opposite legs of the right triangle modeling the geometry of the situation.

The angle relation is ...

tan(angle) = opposite/adjacent

tan(angle) = 550/1050 = 11/21

angle = arctan(11/21) ≈ 27.6°

The angle of elevation is about 27.6°.