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 Answer: 119.4
Units are in miles
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 Answer: 119.4
Units are in miles