Mr. Jameson is walking toward his nine-story hotel and stops to look up to the top of the building. The angle of elevation to the top of the hotel is 40°. If the height of the hotel is 90 feet, about how far away is Mr. Jameson from the hotel?
Accepted Solution
A:
The problem can be modeled as a rectangle triangle where we know a side and an angle. We can use the following trigonometric relationship: tan (x) = (C.O) / (C.A) Where, x: angle C.O: opposite leg C.A: adjoining catheto Substituting values we have: tan (40) = (90) / (C.A) Clearing C.A: C.A = (90) / (tan (40)) C.A = 107.2578233 feet Answer: Mr. Jameson is 107.2578233 feet from the hotel