Question

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?

Answer 1

Mr. Jameson is 107.25 ft far from the hotel.

Explanation:

The arrangement is shown in the figure given.

We need to find value of AB

We have

`tan40=(BC)/(AB)=(90)/(AB)\\\\AB=(90)/(tan40)=107.25ft`

So, Mr. Jameson is 107.25 ft far from the hotel.

Answer 2

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