Geomatics Engineering: 구과량 증명

Question

Girard’s Theorem

A spherical triangle on the surface of a sphere of radius \(r\), with angles \(A, B\) and \(C\), has area, \(T\), given by

\[T = r^{2}(A+B+C-\frac{1}{2}\tau)\text{,}\] \[\text{where }\tau = 2\pi.\]

\[\text{(1) Prove Girard's Theorem.}\] \[\text{(2) Prove that } \varepsilon^{''} = \frac{T}{r^{2}}\rho^{''}.\]

Answer

(1)

\[Proof\] \[\text{A sphere of radius } r \text{ has surface area } 2\tau{r^{2}}.\]

\[\text{If two great circles meet in a lunar angle }\theta, 0 < \theta \leq{\tau}, \text{then the proportion of surface area which is occupied by the lune they create is } \frac{\theta}{\tau}.\] \[\therefore \text{ We have area of lune with lunar angle } \theta.\] \[\frac{\theta}{\tau} \times 2\tau{r^{2}} = 2r^{2}\theta\]

\[\text{Denote by } L_{A}, L_{B} \text{ and } L_{C} \text{ the areas of the three lunes with angles } A, B \text{ and } C, \text{ respectively}.\] \[\text{Also, Denote by } T \text{ the area of triangle } ABC;\]

\[\text{These areas}(L_{A}, L_{B}, L_{C}) \text{ each have an antipodal duplicate.}\] \[\text{Likewise, } T \text{ also has its antipodal duplicate.}\] \[\therefore 2L_{A} + 2L_{B} + 2L_{C} = 2r^{2}\tau + 4T\]

\[T = \frac{1}{2}(L_{A} + L_{B} + L_{C} - r^2\tau)\] \[= \frac{1}{2}(2r^{2}A + 2r^{2}B + 2r^{2}C - r^{2}\tau)\] \[= r^{2}(A + B + C - \frac{1}{2}\tau)\]

(2)

\[Proof\] \[\text{Let's replace } A + B + C - \frac{1}{2}\tau \text{ with } \varepsilon.\] \[\therefore T = r^{2}\varepsilon\] \[\varepsilon = \frac{T}{r^{2}}\]

\[\text{Here,}\] \[\rho^{''} = \frac{180^{\circ}}{\pi}\] \[\therefore \varepsilon^{''} = \frac{T}{r^{2}}\rho^{''}\]