The angle of inclination from the base of skyscraper A to the top of skyscraper B is approximately 10.6degrees. If skyscraper B is 1477 feet tall, how far apart are the two skyscrapers? Assume the bases of the two buildings are at the same elevation.

We can solve this problem drawing a triangle, where one vertex is the base of skyscraper A, the second vertex is the top of skyscraper B, and the third is the base of skyscraper B.

The height of skyscraper B will be the triangle side opposite to the 10.6 angle. So, to know the distance between the skyscraper's bases, we use the tangent of the angle, as it is calculated by the opposite cathetus (1477 feet) over the adjacent cathetus (the distance we want: let's call it "d"):

tangent(10.6) = 1477/d = 0.1871

d = 1477/0.1871 = 7892 feet.

The angle of inclination from the base of skyscraper A to the top of skyscraper B is approximately 10.6degrees. If skyscraper B is 1477 feet​ tall, how far apart are the two​ skyscrapers? Assume the bases of the two buildings are at the same elevation.

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The angle of inclination from the base of skyscraper A to the top of skyscraper B is approximately 10.6degrees. If skyscraper B is 1477 feet tall, how far apart are the two skyscrapers? Assume the bases of the two buildings are at the same elevation.