The excess (A+B+C) - 180 is in fact directly proportional to the spherical area of T. If we measure the angles in radians, then we have the precise assertion:
If A, B, and C are the interior angles of a triangle on a unit sphere, the excess (A+B+C) - Pi is exactly equal to the area of the triangle.
This makes some intuitive sense. For example, a very small triangle will be close to a plane triangle, its area will be small, its excess near 0. An octant triangle with three 90 degree angles will be 1/8 of the sphere, hence of area 4 Pi * (1/8) = Pi / 2, while its defect will be 3 * (Pi / 2) - Pi = Pi/2. I have been told that this result was discovered by the English mathematician Thomas Harriott, in about the year 1620.
I do not know Harriott's proof. The proof which will be given below involves essentially no calculation, but I do not know to whom it should be attributed. Before it is presented, a much simpler fact about areas and angles on a sphere has to be understood. This concerns the area of a sector, the region on a sphere you see when you tear away an orange slice. It is very simple to see that since the total area of a unit sphere is 4 Pi and an angle A is A/2 Pi of a full rotation of 2 Pi:
The area of a sector with vertex angle A (measured in radians) is 4 Pi * A / 2 Pi = 2 A.
For example, a hemisphere is a sector with vertex angle Pi and area 2 Pi.
Now for the original claim, which is proven by the following figure together with the formula for the area of sectors. The point is that each interior angle of a spherical triangle may be extended to a full sector. Keep in mind that if two regions X and Y overlap in the region Z, then the area of the union of X and Y is the sum of the areas of X and Y minus the area of Z.
Hint: Assume A corresponds to the red sector, B to the pink one, C to the magenta one. Let T be the area of the triangle. What is the area of each of the coloured regions? Why do they, together with the two copies of the triangle, cover the sphere?
Finally, both of the assertions above are special cases of a very general claim. In the Euclidean plane, if we wander around the outside of any polygon, then when we get back to our starting point, facing in our starting direction we have turned through a complete circle, or through 360 degrees. On the sphere, the total turning angle of any path will always be different from that. For example, if we wander around a great circle our turning angle is zero degrees. Note that the turning angle is measured strictly on the sphere itself, and that in this sense a great circle never turns, but just goes straight ahead.
For any spherical polygonal figure, define its defect to be the difference between 2 Pi and the total turning angle. It is the sum of the exterior angles of the figure.
The defect of any polygonal figure on a unit sphere is equal to its area. Equivalently, the total turning angle of any polygonal figure on the sphere is equal to 2 Pi minus its area.
The proof uses the result above concerning the sum of angles in a spherical triangle, after decomposing the polygonal figure into triangles and summing things up.
Exercise. Can the turning angle be negative? What does it mean if it is? Explain one simple example.