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:

$$\begin{gathered} A=23+27/60=\frac{469}{20}=23.45{}^{\circ } \\ B=66+33/60=\frac{1331}{20}=66.55{}^{\circ } \\ r=6371\text{km} \\ {h}_1=r-r\cdot \cos (A°\to \text{rad})=r-r\cdot \cos (A°\cdot \frac{\pi }{180})=6371-6371\cdot \cos (23.45°\cdot \frac{3.1415926}{180})=526.19555\text{km} \\ {h}_2=r-r\cdot \cos (B°\to \text{rad})=r-r\cdot \cos (B°\cdot \frac{\pi }{180})=6371-6371\cdot \cos (66.55°\cdot \frac{3.1415926}{180})=3835.66927\text{km} \\ S=4\pi \cdot r^2/2=4\cdot 3.1416\cdot 6371^2/2\doteq 255032235.9549{\text{km}}^2 \\ {S}_1=2\pi \cdot r\cdot h_1=2\cdot 3.1416\cdot 6371\cdot 526.1955\doteq 21063699.1055{\text{km}}^2 \\ {S}_2=2\pi \cdot r\cdot h_2=2\cdot 3.1416\cdot 6371\cdot 3835.6693\doteq 153542506.7173{\text{km}}^2 \\ {p}_1=100\cdot \frac{S_1}{S}=100\cdot \frac{\frac{400210283}{19}}{\frac{5610709191}{22}}=8.2592\% \end{gathered}$$

$$\begin{gathered} S_3=S_2-S_1=\frac{1074797547}{7}-\frac{400210283}{19}=\frac{20421153393}{133}-\frac{2801471981}{133}=\frac{20421153393-2801471981}{133}=\frac{17619681412}{133}\doteq 132478807.6118{\text{km}}^2 \\ {p}_2=100\cdot \frac{S_3}{S}=100\cdot \frac{\frac{662394038}{5}}{\frac{5610709191}{22}}=51.9459\% \end{gathered}$$

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

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