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  "body": "\n## Statement\n\nLet $R \u003e 0$, and let $T$ be a spherical triangle on a sphere of radius $R$, with angles $\\alpha$, $\\beta$, $\\gamma$.\nThen the area of $T$ is\n$$\\operatorname{Area}(T) = R^2 (\\alpha + \\beta + \\gamma - \\pi).$$\n\n## Remarks\n\nThe quantity $E = \\alpha + \\beta + \\gamma - \\pi$ is the *spherical excess*; the theorem shows that it is positive.\nProof idea: a lune, the region between two half great circles meeting at angle $\\theta$, has area $2\\theta R^2$, since it is the fraction $\\theta / (2\\pi)$ of the sphere of area $4\\pi R^2$.\nThe three great circles through the sides of $T$ cut the sphere into eight triangles.\nLet $T'$ be the antipodal triangle; the six lunes with angles $\\alpha$, $\\beta$, $\\gamma$ at the vertices of $T$ and of $T'$ cover the sphere.\nThey cover $T$ and $T'$ three times each and every other point once, so $2(2\\alpha + 2\\beta + 2\\gamma) R^2 = 4\\pi R^2 + 4 \\operatorname{Area}(T)$, which gives the formula.\nExample: the octant triangle with vertices $(R, 0, 0)$, $(0, R, 0)$, $(0, 0, R)$ has three right angles, so its area is $R^2 (3\\pi/2 - \\pi) = \\pi R^2 / 2$, one eighth of $4\\pi R^2$.\n\n## Explanations\n\n### L1 human\n\nOn a sphere, the angles of a triangle add up to more than $180^\\circ$.\nGirard's theorem says that the extra amount, the excess, is exactly the area of the triangle divided by $R^2$.\nSo on a globe one can find the area of a triangle just by measuring its three angles.\n\n### L2 human\n\nThe theorem is the spherical case of the [[Gauss–Bonnet theorem|ESGEHKPD60]]: the integral of the Gaussian curvature $1/R^2$ over the triangle equals the angle excess.\nThe hyperbolic counterpart says that a [[hyperbolic triangle|GCZ8EEDCHS]] in curvature $-1$ has area $\\pi - (\\alpha + \\beta + \\gamma)$, the defect.\nSince the excess is positive and tends to $0$ for small triangles, it explains why the angle sum of a small spherical triangle is close to $\\pi$.\nIt also shows that two spherical triangles with the same angles have the same area.\nIn fact such triangles are congruent, by the law of cosines for angles.\n",
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