Earth has a radius of 3959 miles. a pilot is flying at a steady altitude of 1.8 miles above the earth's surface. what is the pilot's distance to the horizon? (no pic) enter your answer, rounded to the nearest tenth, in the box. mi

This is optional, but I find it easier to solve problems like this if you draw out what it might look like. That's what I've done for the attached image (see below). 

I've defined these points
A = center of earth
B = point just below the plane
C = plane's location
D = horizon (furthest point the pilot can see out)

Based on those point labels, we are given
AB = 3959 miles
BC = 1.8 miles
AD = 3959 miles
CD = x miles

Triangle ADC is a right triangle. So we can use the pythagorean theorem to find x
a^2 + b^2 = c^2
(AD)^2 + (CD)^2 = (AC)^2
3959^2 + x^2 = (AB+BC)^2
3959^2 + x^2 = (3959+1.8)^2
3959^2 + x^2 = 3960.8^2
15,673,681 + x^2 = 15,687,936.64
x^2 = 15,687,936.64 - 15,673,681
x^2 = 14,255.6399999988
x = sqrt(14,255.6399999988)
x = 119.396984886549
x = 119.4 <<--- rounding to the nearest tenth (one decimal place)

Final 119.4
Units are in miles

Bruh sound effect #2

Can u send a screenshot of it? !