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 ° 2 7^{'} = 23 ° + \frac{2 7}{6 0} ° = 23.45 ° = 23.45 B = 66 ° 3 3^{'} = 66 ° + \frac{3 3}{6 0} ° = 66.55 ° = 66.55 r = 6371 km h_{1} = r - r \cdot cos (A ° \to rad) = r - r \cdot cos (A ° \cdot \frac{π}{1 8 0}) = 6371 - 6371 \cdot cos (23.45 ° \cdot \frac{3 . 1 4 1 5 9 2 6}{1 8 0}) = 526.19555 km h_{2} = r - r \cdot cos (B ° \to rad) = r - r \cdot cos (B ° \cdot \frac{π}{1 8 0}) = 6371 - 6371 \cdot cos (66.55 ° \cdot \frac{3 . 1 4 1 5 9 2 6}{1 8 0}) = 3835.66927 km S = 4 π \cdot r^{2} / 2 = 4 \cdot 3.1416 \cdot 637 1^{2} / 2 ≐ 255032235.9549 km^{2} S_{1} = 2 π \cdot r \cdot h_{1} = 2 \cdot 3.1416 \cdot 6371 \cdot 526.1955 ≐ 21063699.1055 km^{2} S_{2} = 2 π \cdot r \cdot h_{2} = 2 \cdot 3.1416 \cdot 6371 \cdot 3835.6693 ≐ 153542506.7173 km^{2} p_{1} = 100 \cdot \frac{S _{1}}{S} = 100 \cdot \frac{\frac{4 0 0 2 1 0 2 8 3}{1 9}}{\frac{5 6 1 0 7 0 9 1 9 1}{2 2}} = 8.2592 %

S_{3} = S_{2} - S_{1} = \frac{1 0 7 4 7 9 7 5 4 7}{7} - \frac{4 0 0 2 1 0 2 8 3}{1 9} = \frac{2 0 4 2 1 1 5 3 3 9 3}{1 3 3} - \frac{2 8 0 1 4 7 1 9 8 1}{1 3 3} = \frac{2 0 4 2 1 1 5 3 3 9 3 - 2 8 0 1 4 7 1 9 8 1}{1 3 3} = \frac{1 7 6 1 9 6 8 1 4 1 2}{1 3 3} ≐ 132478807.6118 km^{2} p_{2} = 100 \cdot \frac{S _{3}}{S} = 100 \cdot \frac{\frac{6 6 2 3 9 4 0 3 8}{5}}{\frac{5 6 1 0 7 0 9 1 9 1}{2 2}} = 51.9459 %

p_{3} = 100 - (p_{1} + p_{2}) = 100 - (8.2592 + 51.9459) = 39.7949 %

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