# 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:

$A=23 + 27/60=\dfrac{ 469 }{ 20 }=23.45 \ ^\circ \ \\ B=66 + 33/60=\dfrac{ 1331 }{ 20 }=66.55 \ ^\circ \ \\ r=6371 \ \text{km} \ \\ \ \\ h_{1}=r - r \cdot \ \cos( A ^\circ \rightarrow\ \text{rad} )=r - r \cdot \ \cos( A ^\circ \cdot \ \dfrac{ \pi }{ 180 } \ )=6371 - 6371 \cdot \ \cos( 23.45 ^\circ \cdot \ \dfrac{ 3.1415926 }{ 180 } \ )=526.19555 \ \\ h_{2}=r - r \cdot \ \cos( B ^\circ \rightarrow\ \text{rad} )=r - r \cdot \ \cos( B ^\circ \cdot \ \dfrac{ \pi }{ 180 } \ )=6371 - 6371 \cdot \ \cos( 66.55 ^\circ \cdot \ \dfrac{ 3.1415926 }{ 180 } \ )=3835.66927 \ \\ \ \\ S=4 \pi \cdot \ r^2 / 2=4 \cdot \ 3.1416 \cdot \ 6371^2 / 2 \doteq 255032235.955 \ \text{km}^2 \ \\ \ \\ S_{1}=2 \pi \cdot \ r \cdot \ h_{1}=2 \cdot \ 3.1416 \cdot \ 6371 \cdot \ 526.1955 \doteq 21063699.1055 \ \text{km}^2 \ \\ S_{2}=2 \pi \cdot \ r \cdot \ h_{2}=2 \cdot \ 3.1416 \cdot \ 6371 \cdot \ 3835.6693 \doteq 153542506.717 \ \text{km}^2 \ \\ \ \\ p_{1}=100 \cdot \ \dfrac{ S_{1} }{ S }=100 \cdot \ \dfrac{ 21063699.1055 }{ 255032235.955 }=8.259 \%$
$S_{3}=S_{2} - S_{1}=153542506.717 - 21063699.1055=132478807.611 \ \text{km}^2 \ \\ p_{2}=100 \cdot \ \dfrac{ S_{3} }{ S }=100 \cdot \ \dfrac{ 132478807.611 }{ 255032235.955 }=51.946 \%$
$p_{3}=100 - (p_{1}+p_{2})=100 - (8.2592+51.9459)=\dfrac{ 7959 }{ 200 }=39.795 \%$

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