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

Answer:

Mr. Jameson is 107.25 ft far from the hotel.

Step-by-step explanation:

The arrangement is shown in the figure given.

We need to find value of AB

We have

            $tan40=\frac{BC}{AB}=\frac{90}{AB}\\\\AB=\frac{90}{tan40}=107.25ft$

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