An airplane rises vertically 1000 feet over a horizontal distance of 1 mile. What is the angle of elevation of the airplane's path? (hint: 1 mile = 5280 feet)

I solved this one under a previous post.

To find the angle of elevation of the airplane's path, we can use trigonometry.

First, let's convert the horizontal distance of 1 mile to feet. Given that 1 mile is equal to 5280 feet, we can multiply the distance by 5280 to get the total distance in feet:

1 mile * 5280 feet/mile = 5280 feet

Now we have the vertical distance of 1000 feet and the horizontal distance of 5280 feet. To find the angle of elevation, we can use the tangent function since tangent is defined as the ratio of the opposite side (vertical distance) to the adjacent side (horizontal distance):

tan(angle) = opposite / adjacent

Substituting the values we have:

tan(angle) = 1000 feet / 5280 feet

Now, we can solve for the angle by taking the inverse tangent (arctan) of both sides of the equation:

angle = arctan(1000 feet / 5280 feet)

Using a calculator or computer software that has the inverse tangent function, we find that:

angle ≈ 10.6 degrees

Therefore, the angle of elevation of the airplane's path is approximately 10.6 degrees.