# Tropics and polar zones

What percentage of the Earth’s surface lies in the tropical, temperate and polar zone?
Individual zones are bordered by tropics 23°27' and polar circles 66°33'

Result

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

#### Solution:

$T=23 + 27/60=\dfrac{ 469 }{ 20 }=23.45 \ ^\circ \ \\ P=66 + 33/60=\dfrac{ 1331 }{ 20 }=66.55 \ ^\circ \ \\ \ \\ r=1 \ \\ \cos (90-P)=x_{1} : r \ \\ x_{1}=r \cdot \ \cos( ( 90-P) ^\circ \rightarrow\ \text{rad})=r \cdot \ \cos( ( 90-P )^\circ \cdot \ \dfrac{ \pi }{ 180 } \ )=1 \cdot \ \cos( ( 90-66.55 )^\circ \cdot \ \dfrac{ 3.1415926 }{ 180 } \ )=0.91741 \ \\ h_{1}=r - x_{1}=1 - 0.9174 \doteq 0.0826 \ \\ \ \\ \cos (90-T)=x_{2} : r \ \\ x_{2}=r \cdot \ \cos( ( 90-T) ^\circ \rightarrow\ \text{rad})=r \cdot \ \cos( ( 90-T )^\circ \cdot \ \dfrac{ \pi }{ 180 } \ )=1 \cdot \ \cos( ( 90-23.45 )^\circ \cdot \ \dfrac{ 3.1415926 }{ 180 } \ )=0.39795 \ \\ h_{2}=r - x_{2}=1 - 0.3979 \doteq 0.6021 \ \\ \ \\ S_{1}=2 \pi \cdot \ r \cdot \ h_{1}=2 \cdot \ 3.1416 \cdot \ 1 \cdot \ 0.0826 \doteq 0.5189 \ \\ S_{2}=2 \pi \cdot \ r \cdot \ h_{2}=2 \cdot \ 3.1416 \cdot \ 1 \cdot \ 0.6021 \doteq 3.7828 \ \\ \ \\ S_{3}=2 \pi \cdot \ r^2=2 \cdot \ 3.1416 \cdot \ 1^2 \doteq 6.2832 \ \\ \ \\ p_{1}=100 \cdot \ \dfrac{ S_{3}-S_{2} }{ S_{3} }=100 \cdot \ \dfrac{ 6.2832-3.7828 }{ 6.2832 } \doteq 39.7949 \doteq 39.795 \%$
$p_{2}=100 \cdot \ \dfrac{ S_{2}-S_{1} }{ S_{3} }=100 \cdot \ \dfrac{ 3.7828-0.5189 }{ 6.2832 } \doteq 51.9459 \doteq 51.946 \%$
$p_{3}=100 \cdot \ \dfrac{ S_{1} }{ S_{3} }=100 \cdot \ \dfrac{ 0.5189 }{ 6.2832 } \doteq 8.2592 \doteq 8.259 \%$

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