Tropical, mild and arctic

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

Correct answer:

p₁ =  8.2592 %
p₂ =  51.9459 %
p₃ =  39.7949 %

Step-by-step explanation:

$A = 23\degree 27 ' = 23 \degree + \dfrac{ 27 }{ 60 } \degree = 23.45 \degree = 23.45 \ \\ B = 66\degree 33 ' = 66 \degree + \dfrac{ 33 }{ 60 } \degree = 66.55 \degree = 66.55 \ \\ r = 6371 \ \text{km} \ \\ \ \\ h_{1} = r - r \cdot \ \cos( A \degree \rightarrow\ \text{rad} ) = r - r \cdot \ \cos( A \degree \cdot \ \dfrac{ \pi }{ 180 } \ ) = 6371 - 6371 \cdot \ \cos( 23.45 \degree \cdot \ \dfrac{ 3.1415926 }{ 180 } \ ) = 526.19555 \ \text{km} \ \\ h_{2} = r - r \cdot \ \cos( B \degree \rightarrow\ \text{rad} ) = r - r \cdot \ \cos( B \degree \cdot \ \dfrac{ \pi }{ 180 } \ ) = 6371 - 6371 \cdot \ \cos( 66.55 \degree \cdot \ \dfrac{ 3.1415926 }{ 180 } \ ) = 3835.66927 \ \text{km} \ \\ \ \\ S = 4 \pi \cdot \ r^2 / 2 = 4 \cdot \ 3.1416 \cdot \ 6371^2 / 2 \doteq 255032235.9549 \ \text{km}^2 \ \\ \ \\ S_{1} = 2 \pi \cdot \ r \cdot \ h_{1} = 2 \cdot \ 3.1416 \cdot \ 6371 \cdot \ 526.1955 \doteq 21063699.1055 \ \text{km}^2 \ \\ S_{2} = 2 \pi \cdot \ r \cdot \ h_{2} = 2 \cdot \ 3.1416 \cdot \ 6371 \cdot \ 3835.6693 \doteq 153542506.7173 \ \text{km}^2 \ \\ \ \\ p_{1} = 100 \cdot \ \dfrac{ S_{1} }{ S } = 100 \cdot \ \dfrac{ 21063699.1055 }{ 255032235.9549 } = 8.2592 \%$

$S_{3} = S_{2} - S_{1} = 153542506.7173 - 21063699.1055 \doteq 132478807.6118 \ \text{km}^2 \ \\ p_{2} = 100 \cdot \ \dfrac{ S_{3} }{ S } = 100 \cdot \ \dfrac{ 132478807.6118 }{ 255032235.9549 } = 51.9459 \%$

$p_3=100-(p_1+p_2)=100-(8.2592+51.9459)=39.7949\%$

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