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'.

Final Answer:

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

Step-by-step explanation:

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

$$\begin{gathered} S_3=S_2-S_1=153542506.7173-21063699.1055\doteq 132478807.6118{\text{km}}^2 \\ {p}_2=100\cdot \frac{S_3}{S}=100\cdot \frac{132478807.6118}{255032235.9549}=51.9459\% \end{gathered}$$

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

Try another example