How can spherical triangle be defined in this way? Let $u=\frac1{\sqrt2}(1,-1,0)$, $v=\frac{1}{\sqrt2}(1,1,0)$, $w=\frac1{\sqrt2}(0,1,1)$. I want to find the area of the spherical triangle $T=\left\{x\in S^2:d_S(x,u)\le \frac{\pi}2, d_S(x,v)\le \frac{\pi}2,d_S(x,w)\le\frac{\pi}2\right\}$.
I'm wondering what does this definition of a spherical triangle mean? I think $d_S(x,u)$ is a spherical disc of radius $\frac\pi2$, so does this definition mean that we have three spherical discs of this radius around $u,v$ and $w$? But how does this make a triangle?
 A: What may be initially confusing about this question is that we often see
a spherical triangle described by the coordinates of its three vertices.
But in this question, the points $u, v, w$ are not the vertices of the triangle. They are poles of the great circles that form the three sides of the triangle.
The text of the question says $T$ is a triangle, but the mathematical notation actually defines it as the intersection of three hemispherical regions. A region like $\left\{x \mid d_S(x,u) \le \delta\right\}$ is a spherical cap centered at $u$; if we make this cap large enough it will cover exactly half of the sphere, and this happens when $\delta = \frac\pi2.$
So the region $U = \left\{x \mid d_S(x,u) \leq \frac\pi2\right\}$
is a hemisphere. The boundary of the hemisphere is the great circle
$\partial U = \left\{x \mid d_S(x,u) = \frac\pi2\right\},$
and the interior of the hemisphere consists of all points on the same side of
$\partial U$ as $u.$
To find $T,$ we intersect the hemisphere $U$ with the hemispheres
$V = \left\{x \mid d_S(x,v) \leq \frac\pi2\right\}$
and $W = \left\{x \mid d_S(x,w) \leq \frac\pi2\right\}.$
The intersection of $U$ and $V$ is a spherical lune bounded by semicircles belonging to the great circles $\partial U$ and $\partial V.$
The great circle $\partial W$ intersects $\partial U$ at two antipodal points, and since neither of those is at the intersections of
$\partial U$ and $\partial V$ (all three poles would have to lie on one great circle for this to happen), $\partial W$ cuts across the semicircular part of 
$\partial U$ that bounds the lune, and cuts the lune in two triangular pieces.
One of those triangles lies in the hemisphere $W,$ and that triangle is exactly the intersection of the three hemispheres $U,$ $V,$ and $W,$
so it is the region $T.$
Another way to look at it is that the great circle
$\partial U = \left\{x \mid d_S(x,u) = \frac\pi2\right\},$
the great circle
$\partial V = \left\{x \mid d_S(x,v) = \frac\pi2\right\},$
and the great circle 
$\partial W = \left\{x \mid d_S(x,w) = \frac\pi2\right\}$
divide the sphere into eight triangular regions like a Venn diagram of three sets.
As in a Venn diagram, the intersection of the three sets is the region
that is "inside" all three circles, that is, it is on the same side of
the great circle $\partial U$ as $u,$
on the same side of $\partial V$ as $v,$ and on the same side of $\partial W$ as $w.$
The vertices of the triangle are at intersections of the great circles because that's where the part of the boundary of $T$ contributed by one great circle meets the part of the boundary of $T$ contributed by another great circle.
For example, the great circles $\partial U$ and $\partial V$ intersect at two points.
One of these two points (the one closer to $w$) is a vertex of the
desired spherical triangle. Call this point $A.$
Since we are on the unit sphere, 
the two great circles $\partial U$ and $\partial V$ make an angle at $A$ 
that is equal to the arc distance from $u$ to $v.$
Using this information you can find the angle of the vertex $A.$
You can find the angles at the other two vertices by finding the arc distances from $u$ to $w$ and $v$ to $w.$
Since you can derive the area of a spherical triangle from just the information about the angles at its three vertices, you do not actually need to find all three vertices of the triangle, just the distances between the three poles.
A: The boundary of the triangle is formed by arcs of circumference of $\pi/2$ of amplitude. You say you wnt to understand this.
$T=\left\{x\in S^2:d_S(x,u)\le \frac{\pi}2, d_S(x,v)\le \frac{\pi}2,d_S(x,w)\le\frac{\pi}2\right\}$
To see intuitively what this region is, draw it in the plane and then try to apply to the sphere:

Some points first to consider about the drawing: the point $D$ is interior to the region. We can check that $d_P(D,V_i)\lt 4$ for $V_i=A,B,C$. Consider now $E$, that has $d_P(B,E)\lt4$ and $d_P(C,E)\lt4$, but $d_P(A,E)\gt4$ and it's in the exterior. So, the triple condition for a point of having $d_P(D,V_i)\lt 4$ determines the interior of the region.
You can apply the image to your problem with some easy changes. First, $x$ has to be the position vector for some point in $S^2$. Second $d_S$ stands for the distance along $S^2$. We need to see that the arcs forming the boundary of the region, drawn this time along the sphere, are maximum arcs, so is, "straight" lines for the sphere. To see this, consider one of the points, say $u$, and set it as the north pole. Then the points at a distance (along a sphere of radius $1$) of $\pi/2$ are the equatorial ones, so is, they are on a maximum arc. Then, a piece of equator is in the boundary of $T$. But it is this way no mind wich of the points $u\;,v\;,w$ we can choose. All the arcs are maximum arcs: the region is a spherical triangle.
The area is easy to calculate considering that a sphere can be "patched" with copies of this region.
