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=rr cos(A rad)=rr cos(A π180 )=63716371 cos(23.45 3.1415926180 )=526.19555 h2=rr cos(B rad)=rr cos(B π180 )=63716371 cos(66.55 3.1415926180 )=3835.66927  S=4π r2/2=4 3.1416 63712/2255032235.955 km2  S1=2π r h1=2 3.1416 6371 526.195521063699.1055 km2 S2=2π r h2=2 3.1416 6371 3835.6693153542506.717 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=S2S1=153542506.71721063699.1055=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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