MATH SOLVE

10 months ago

Q:
# Cesar is building a shelf at an angle so that it appears to be leaning against the wall. The shelf touches the wall 7 ft. above the floor and the distance between the floor from the wall to the front of the shelf is 1.5 ft. What is the angle the shelf makes with the floor?A. 77.9 degrees. B. 77.6 degrees.C. 12.1 degrees.D. 12.4 degrees. A 15-ft. building casts a shadow of 12 ft. What is the approximate angle of inclination of the sun? A. 51.3 degrees. B. 53.1 degrees. C. 36.9 degrees. D. 19.2 degrees.

Accepted Solution

A:

1) The problem says:

- The shelf touches the wall 7 feet above the floor.

-The distance between the floor from the wall to the front of the shelf is 1.5 feet

Then:

Tan^-1(α)=Opposite Leg/Adjacent leg

Opposite leg=7

Adjacent leg=1.5

When you substitute the values, you obtain:

Tan^-1(α)=7/1.5

α=77.9°

What is the angle the shelf makes with the floor?

The correct answer is: A) 77.9 degrees.

2) The problem says that the 15-feet building casts a shadow of 12 feet. Then:

Tan^-1(β)=Opposite Leg/Adjacent leg

Opposite leg=15

Adjacent leg=12

When you substitute, you obtain:

Tan^-1(β)=15/12

β=51.3°

What is the approximate angle of inclination of the sun?

The correct answer is: A) 51.3 degrees.

- The shelf touches the wall 7 feet above the floor.

-The distance between the floor from the wall to the front of the shelf is 1.5 feet

Then:

Tan^-1(α)=Opposite Leg/Adjacent leg

Opposite leg=7

Adjacent leg=1.5

When you substitute the values, you obtain:

Tan^-1(α)=7/1.5

α=77.9°

What is the angle the shelf makes with the floor?

The correct answer is: A) 77.9 degrees.

2) The problem says that the 15-feet building casts a shadow of 12 feet. Then:

Tan^-1(β)=Opposite Leg/Adjacent leg

Opposite leg=15

Adjacent leg=12

When you substitute, you obtain:

Tan^-1(β)=15/12

β=51.3°

What is the approximate angle of inclination of the sun?

The correct answer is: A) 51.3 degrees.