Angles and area of a triangle defined on a sphere? 
Consider the spherical triangle $\mathcal{P}$ with vertices $P_1 = (1,0,0)$, $P_2 = (0,1,0)$ and $P_3 = (1/\sqrt{3}, 1/\sqrt{3},1/\sqrt{3})$. Find the angles $\phi_1, \phi_2, \phi_3$ of $\mathcal{P}$ at $P_1, P_2, P_3$ respectively. 

My first question: what's the relation between cartesian to spherical coordinate change  $(x,y, z)$ $= (r\sin \theta \cos \phi, r \sin\theta \sin \phi, r \cos \theta)$ and angles of the spherical triangle? 
How do I find $\phi_1, \phi_2, \phi_3$? I've calculated $\theta$ from $r\cos{\theta} = 0$ which resulted in $\theta = \pi/2$. 
But I suspect my $\theta$ and $\phi$ may not be related to $\phi_1, \phi_2, \phi_3$. Is this the case? Thank you. 
I think I know how to find the area after I've found these angles (regarding the title). 
 A: As mentioned in my comment, the specific points used in this problem make various values "obvious". However, I'll crunch through a computational solution without leveraging that obvious-ness (but see below), because it's good to have ways to proceed when obvious things don't seem so obvious.

To recap a bit of spherical geometry ... Side $P_2P_3$ of the spherical triangle $\triangle P_1P_2P_3$ is the great circle arc subtending central angle $\angle P_2OP_3$. Since the sphere (and the great circle) has unit radius, arc-length and central-angle-measure are interchangeable; therefore, we can calculate the cosines of the sides of the triangle with dot products. (In what follows, I define $p_1 := |P_2P_3|$, etc.)
$$\begin{align}
\cos p_1 = \cos\angle P_2OP_3 = P_2\cdot P_3 &= 1/\sqrt{3} = \cos p_2 \\[4pt]
\cos p_3 = \cos\angle P_1OP_2 = P_1\cdot P_2 &= 0
\end{align}$$
Since $\sin^2\theta + \cos^2\theta = 1$ (and "knowing" that sine is non-negative for the angles in question), we may calculate
$$\sin p_1 = \sin p_2 = \sqrt{1-\left(\frac{1}{\sqrt{3}}\right)^2} = \frac{\sqrt{2}}{\sqrt{3}} \qquad\qquad \sin p_3 = 1$$
From here, we leverage the spherical law of cosines. For instance,
$$\cos\phi_1 = \frac{\cos p_1 - \cos p_2\cos p_3}{\sin p_2 \sin p_3} = 
\frac{\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{3}}\cdot 0}{\frac{\sqrt{2}}{\sqrt{3}}\cdot 1} = \frac{1}{\sqrt2}$$
$$\cos\phi_2 = \frac{\cos p_2 - \cos p_3\cos p_1}{\sin p_3 \sin p_1} = 
\frac{\frac{1}{\sqrt{3}}-0\cdot\frac{1}{\sqrt{3}}}{1\cdot\frac{\sqrt{2}}{\sqrt{3}}} = \frac{1}{\sqrt2}$$
$$\cos\phi_3 = \frac{\cos p_3 - \cos p_1\cos p_2}{\sin p_1 \sin p_2} = 
\frac{0-\frac{1}{\sqrt{3}}\cdot\frac{1}{\sqrt{3}}}{\frac{\sqrt{2}}{\sqrt{3}}\cdot \frac{\sqrt{2}}{\sqrt{3}}} = -\frac12$$
From here, we see that $\phi_1$, $\phi_2$, $\phi_3$ are the "special" angles $\frac14\pi$, $\frac14\pi$, $\frac23\pi$. (The reader can compute the triangle's area from these.) It's no coincidence that two of these are half-right angles, while the last is one-third of a full turn; this is, in fact, a consequence of the "obvious-ness" I ignored from the beginning.
As hinted in @MichealBehrend's comment, if we introduce a fourth point ---I'll call it $P_4 = (0,0,1)$--- then $P_1$, $P_2$, $P_4$ are the vertices of an equilateral spherical triangle that fills a full octant of the sphere; each angle of this spherical triangle has measure $\frac12 \pi$ (a fact we'll use shortly). The point $P_3$, as with any point of the form $(k,k,k)$, lies on the line through the origin and the "center" of that octant, so that $P_3$ is the "center" of the triangle.


*

*By three-fold rotational symmetry, we expect the angles formed by joining $P_3$ to the vertices of the triangle to be congruent; thus, each of them (including $\phi_3$) is one-third of a full turn.

*Reflective symmetry implies that center $P_3$ should lie on the angle bisectors at $P_1$, $P_2$, $P_4$. Thus, for instance, $\phi_1 := \angle P_3P_1P_2$ should be half of $\angle P_4P_1P_2$, but that latter angle is a right angle; consequently, $\phi_1 = \frac14\pi$. (Likewise for $\phi_2$.)
This gives the various angles, from which the area of $\triangle P_1P_2P_3$ can be computed. However, there, too, is an "obvious" solution: that triangle is one-third of the full-octant triangle $\triangle P_1P_2P_4$, and the full-octant triangle has measure $4\pi/8 = \pi/2$. Therefore, $|\triangle P_1P_2P_3|= \pi/6$, which agrees with the "spherical excess" calculation $\phi_1+\phi_2+\phi_3-\pi$.
