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The angle of inclination from the base of skyscraper A to the top of skyscraper B is approximately 13.713.7degrees°. If skyscraper B is 14521452 feet​ tall, how far apart are the two​ skyscrapers? Assume the bases of the two buildings are at the same elevation.

1 Answer

6 votes

Answer:

5956 ft

Explanation:

If we designate the points at the bases of the skyscrapers as A and B, and the top of skyscraper B as point T, then triangle ABT is a right triangle with angle B = 90° and angle A = 13.7°.

The definition of the tangent function tells us ...

tan(A) = BT/AB

Solving for AB, we find ...

AB = BT/tan(A) = (1452 ft)/tan(13.7°)

AB ≈ 5956 ft

The skyscrapers are about 5956 feet apart.

User Oleg Apanovich
by
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