Tropics and polar zones

What percentage of the Earth's surface lies in the tropical, temperate, and polar zones?
Tropics border individual zones at 23°27' and polar circles at 66°33'.

Final Answer:

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

Step-by-step explanation:

$$\begin{gathered} T=23+27/60=\frac{469}{20}=23.45{}^{\circ } \\ P=66+33/60=\frac{1331}{20}=66.55{}^{\circ } \\ r=1 \\ \cos (90-P)=x_1:r \\ {x}_1=r\cdot \cos ((90-P)°\to \text{rad})=r\cdot \cos ((90-P)°\cdot \frac{\pi }{180})=1\cdot \cos ((90-66.55)°\cdot \frac{3.1415926}{180})=0.91741 \\ {h}_1=r-x_1=1-0.9174\doteq 0.0826 \\ \cos (90-T)=x_2:r \\ {x}_2=r\cdot \cos ((90-T)°\to \text{rad})=r\cdot \cos ((90-T)°\cdot \frac{\pi }{180})=1\cdot \cos ((90-23.45)°\cdot \frac{3.1415926}{180})=0.39795 \\ {h}_2=r-x_2=1-0.3979\doteq 0.6021 \\ {S}_1=2\pi \cdot r\cdot h_1=2\cdot 3.1416\cdot 1\cdot 0.0826\doteq 0.5189 \\ {S}_2=2\pi \cdot r\cdot h_2=2\cdot 3.1416\cdot 1\cdot 0.6021\doteq 3.7828 \\ {S}_3=2\pi \cdot r^2=2\cdot 3.1416\cdot 1^2\doteq 6.2832 \\ {p}_1=100\cdot \frac{S_3-S_2}{S_3}=100\cdot \frac{6.2832-3.7828}{6.2832}=39.7949\% \end{gathered}$$

$p_2=100\cdot \frac{S_2-S_1}{S_3}=100\cdot \frac{3.7828-0.5189}{6.2832}=51.9459\%$

$p_3=100\cdot \frac{S_1}{S_3}=100\cdot \frac{0.5189}{6.2832}=8.2592\%$

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