Find the area of a spherical triangle made by the points $(0, 0, 1)$, $(0, 1, 0)$ and $(\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}})$. 
Calculate the area of the spherical triangle defined by the points $(0, 0, 1)$, $(0, 1, 0)$ and $(\dfrac{1}{\sqrt{2}}, 0, \dfrac{1}{\sqrt{2}})$.

I have come up with this:
From the spherical Gauss-Bonnet Formula, where $T$ is a triangle with interior angles $\alpha, \beta, \gamma$. Then the area of the triangle $T$ is $\alpha + \beta + \gamma - \pi$.
How do I work out the interior angles in order to use this formula?
Any help appreciated.

$A(0, 0, 1)$, $B(0, 1, 0)$ and $C(\dfrac{1}{\sqrt{2}}, 0, \dfrac{1}{\sqrt{2}})$ with $|A|=|B|=|C|=1$ these point lie on unit sphere. These points specify three plane $x=0$, $y=0$ and $x=z$ then the angle between them are $\dfrac{\pi}{2}$, $\dfrac{\pi}{2}$ and $\dfrac{\pi}{4}$, since their normal vectors are $\vec{i}$, $\vec{j}$ and $\vec{i}-\vec{k}$, respectively (by $\cos\theta=\dfrac{u.v}{|u||v|}=u.v$).
At last $\sigma=\dfrac{\pi}{4}+\dfrac{\pi}{2}+\dfrac{\pi}{2}-\pi=\dfrac{\pi}{4}$.
