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# 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

$T = r^{2}(A+B+C-\frac{1}{2}\tau)\text{,}$ $\text{where }\tau = 2\pi.$ $\text{(1) Prove Girard's Theorem.}$ $\text{(2) Prove that } \varepsilon^{''} = \frac{T}{r^{2}}\rho^{''}.$

(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)

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

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

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