# 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:

$T=23 + 27/60=\dfrac{ 469 }{ 20 }=23.45 \ ^\circ \ \\ P=66 + 33/60=\dfrac{ 1331 }{ 20 }=66.55 \ ^\circ \ \\ \ \\ r=1 \ \\ \cos (90-P)=x_{1} : r \ \\ x_{1}=r \cdot \ \cos( ( 90-P) ^\circ \rightarrow\ \text{rad})=r \cdot \ \cos( ( 90-P )^\circ \cdot \ \dfrac{ \pi }{ 180 } \ )=1 \cdot \ \cos( ( 90-66.55 )^\circ \cdot \ \dfrac{ 3.1415926 }{ 180 } \ )=0.91741 \ \\ h_{1}=r - x_{1}=1 - 0.9174 \doteq 0.0826 \ \\ \ \\ \cos (90-T)=x_{2} : r \ \\ x_{2}=r \cdot \ \cos( ( 90-T) ^\circ \rightarrow\ \text{rad})=r \cdot \ \cos( ( 90-T )^\circ \cdot \ \dfrac{ \pi }{ 180 } \ )=1 \cdot \ \cos( ( 90-23.45 )^\circ \cdot \ \dfrac{ 3.1415926 }{ 180 } \ )=0.39795 \ \\ h_{2}=r - x_{2}=1 - 0.3979 \doteq 0.6021 \ \\ \ \\ S_{1}=2 \pi \cdot \ r \cdot \ h_{1}=2 \cdot \ 3.1416 \cdot \ 1 \cdot \ 0.0826 \doteq 0.5189 \ \\ S_{2}=2 \pi \cdot \ r \cdot \ h_{2}=2 \cdot \ 3.1416 \cdot \ 1 \cdot \ 0.6021 \doteq 3.7828 \ \\ \ \\ S_{3}=2 \pi \cdot \ r^2=2 \cdot \ 3.1416 \cdot \ 1^2 \doteq 6.2832 \ \\ \ \\ p_{1}=100 \cdot \ \dfrac{ S_{3}-S_{2} }{ S_{3} }=100 \cdot \ \dfrac{ 6.2832-3.7828 }{ 6.2832 } \doteq 39.7949 \doteq 39.795 \%$
$p_{2}=100 \cdot \ \dfrac{ S_{2}-S_{1} }{ S_{3} }=100 \cdot \ \dfrac{ 3.7828-0.5189 }{ 6.2832 } \doteq 51.9459 \doteq 51.946 \%$
$p_{3}=100 \cdot \ \dfrac{ S_{1} }{ S_{3} }=100 \cdot \ \dfrac{ 0.5189 }{ 6.2832 } \doteq 8.2592 \doteq 8.259 \%$ Our examples were largely sent or created by pupils and students themselves. Therefore, we would be pleased if you could send us any errors you found, spelling mistakes, or rephasing the example. Thank you!

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