MATH SOLVE

4 months ago

Q:
# please help... The radius of the planet is 4100 miles. Find the distance d to the horizon that a person can see on a clear day from a height of 1298 feet above the planet. (hint: use the conversion 1 mile = 5280 ft) Round your answer to the nearest 10th, and don't forget to convert feet to miles.

Accepted Solution

A:

First, we are going to convert 1298 feet to miles. Since we know that 1 mile = 5280 ft, we are going to multiply 1298 ft by [tex] \frac{1mi}{5280ft} [/tex] to do that:

[tex]1298ft*\frac{1mi}{5280ft}= \frac{59}{240} mi[/tex]

Next, we are going to set up a right triangle using the radius of the Earth and the distance above the planet.

The hypotenuse of our triangle will be the radius of the Earth + the distance above the planet. One of the legs of our triangle will be the radius of the Earth, and the other leg, [tex]d[/tex], will be the distance that a person will see on a clear day.

Using the Pythagorean theorem:

[tex]d^2= (\frac{984059}{240} )^2-4100^2[/tex]

[tex]d= \sqrt{(\frac{984059}{240} )^2-4100^2} [/tex]

[tex]d=44.9[/tex] miles

We can conclude that the distance, [tex]d[/tex], to the horizon that a person can see on a clear day from a height of 1298 feet above the planet is 44.9 miles.

[tex]1298ft*\frac{1mi}{5280ft}= \frac{59}{240} mi[/tex]

Next, we are going to set up a right triangle using the radius of the Earth and the distance above the planet.

The hypotenuse of our triangle will be the radius of the Earth + the distance above the planet. One of the legs of our triangle will be the radius of the Earth, and the other leg, [tex]d[/tex], will be the distance that a person will see on a clear day.

Using the Pythagorean theorem:

[tex]d^2= (\frac{984059}{240} )^2-4100^2[/tex]

[tex]d= \sqrt{(\frac{984059}{240} )^2-4100^2} [/tex]

[tex]d=44.9[/tex] miles

We can conclude that the distance, [tex]d[/tex], to the horizon that a person can see on a clear day from a height of 1298 feet above the planet is 44.9 miles.