Tropical, mild and arctic

How many percents of the Earth's surface lies in the tropical, mild and arctic range? The border between the ranges is the parallel 23°27 'and 66°33'.

Correct result:

p1 =  8.259 %
p2 =  51.946 %
p3 =  39.795 %

Solution:

A=23+27/60=46920=23.45 ∘ B=66+33/60=133120=66.55 ∘ r=6371 km  h1=r−r⋅ cos⁡(A∘→ rad)=r−r⋅ cos⁡(A∘⋅ π180 )=6371−6371⋅ cos⁡(23.45∘⋅ 3.1415926180 )=526.19555 h2=r−r⋅ cos⁡(B∘→ rad)=r−r⋅ cos⁡(B∘⋅ π180 )=6371−6371⋅ cos⁡(66.55∘⋅ 3.1415926180 )=3835.66927  S=4π⋅ r2/2=4⋅ 3.1416⋅ 63712/2≐255032235.955 km2  S1=2π⋅ r⋅ h1=2⋅ 3.1416⋅ 6371⋅ 526.1955≐21063699.1055 km2 S2=2π⋅ r⋅ h2=2⋅ 3.1416⋅ 6371⋅ 3835.6693≐153542506.717 km2  p1=100⋅ S1S=100⋅ 21063699.1055255032235.955=8.259%A=23 + 27/60=\dfrac{ 469 }{ 20 }=23.45 \ ^\circ \ \\ B=66 + 33/60=\dfrac{ 1331 }{ 20 }=66.55 \ ^\circ \ \\ r=6371 \ \text{km} \ \\ \ \\ h_{1}=r - r \cdot \ \cos( A ^\circ \rightarrow\ \text{rad} )=r - r \cdot \ \cos( A ^\circ \cdot \ \dfrac{ \pi }{ 180 } \ )=6371 - 6371 \cdot \ \cos( 23.45 ^\circ \cdot \ \dfrac{ 3.1415926 }{ 180 } \ )=526.19555 \ \\ h_{2}=r - r \cdot \ \cos( B ^\circ \rightarrow\ \text{rad} )=r - r \cdot \ \cos( B ^\circ \cdot \ \dfrac{ \pi }{ 180 } \ )=6371 - 6371 \cdot \ \cos( 66.55 ^\circ \cdot \ \dfrac{ 3.1415926 }{ 180 } \ )=3835.66927 \ \\ \ \\ S=4 \pi \cdot \ r^2 / 2=4 \cdot \ 3.1416 \cdot \ 6371^2 / 2 \doteq 255032235.955 \ \text{km}^2 \ \\ \ \\ S_{1}=2 \pi \cdot \ r \cdot \ h_{1}=2 \cdot \ 3.1416 \cdot \ 6371 \cdot \ 526.1955 \doteq 21063699.1055 \ \text{km}^2 \ \\ S_{2}=2 \pi \cdot \ r \cdot \ h_{2}=2 \cdot \ 3.1416 \cdot \ 6371 \cdot \ 3835.6693 \doteq 153542506.717 \ \text{km}^2 \ \\ \ \\ p_{1}=100 \cdot \ \dfrac{ S_{1} }{ S }=100 \cdot \ \dfrac{ 21063699.1055 }{ 255032235.955 }=8.259 \%
S3=S2−S1=153542506.717−21063699.1055=132478807.611 km2 p2=100⋅ S3S=100⋅ 132478807.611255032235.955=51.946%S_{3}=S_{2} - S_{1}=153542506.717 - 21063699.1055=132478807.611 \ \text{km}^2 \ \\ p_{2}=100 \cdot \ \dfrac{ S_{3} }{ S }=100 \cdot \ \dfrac{ 132478807.611 }{ 255032235.955 }=51.946 \%
p3=100−(p1+p2)=100−(8.2592+51.9459)=7959200=39.795%p_{3}=100 - (p_{1}+p_{2})=100 - (8.2592+51.9459)=\dfrac{ 7959 }{ 200 }=39.795 \%

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