MATH SOLVE

4 months ago

Q:
# 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

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