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.
On
Preliminary remark: (see figure below) It is visible that the area of spherical triangle $ABC$ is a half of the area of spherical triangle $ABD$, whose area is the eighth part of the area of the sphere with radius $r=1$ ( triangle $ABD$ is the part of the sphere situated in the positive orthant). Thus, the final result is:
$$\dfrac{1}{16}(4\pi r^2)=\dfrac{\pi}{4}.$$
As the objective is to use Gauss-Bonnet formula, more precisely Girard's theorem (http://www.princeton.edu/~rvdb/WebGL/GirardThmProof.html), instead of using rather opaque spherical trigonometry formulas, it is clearer IMHO to use (double) cross-products. Here is how:
Let $\vec{A}=(0,0,1), \vec{B}=(0,1,0), \vec{C}=(a,0,a)$ with $a=\dfrac{1}{\sqrt{2}}$.
all of them being on the unit sphere.
Cross product $\vec{C_1}=\vec{A} \times \vec{B}=(-1,0,0)$ is a normal vector to plane $OAB$.
In the same way, cross product $\vec{A_1}=\vec{B} \times \vec{C}=(a,0,-a)$ is a normal vector to plane $OBC$.
Thus, cross product $\vec{C_1} \times \vec{A_1}$ is equal to $(0,-a,0)$ with norm $\dfrac{1}{\sqrt{2}}$ is equal to $\|C_1\|\|A_1\|\sin(\pi-\beta)=1 * 1 * \sin(\beta)$.
Thus $\sin(\beta)=\dfrac{1}{\sqrt{2}}$, whence $\beta=\dfrac{\pi}{4}$.
Remark: in fact, $\beta=3\dfrac{\pi}{4}$ is forbidden because points $A,B,C$ are situated in the positive orthant. Had the elimination of ambiguity not been possible between $\dfrac{\pi}{4}$ and $3\dfrac{\pi}{4}$, we would have used the value of $\cos(\beta)$ obtained using the dot product $\vec{C_1}.\vec{A_1}=\|C_1\|\|A_1\|\cos(\beta)$.
The same process can be used to find angles $\alpha=\dfrac{\pi}{2}$ and $\gamma=\dfrac{\pi}{2}$ (see figure), and then use Girard's formula.
(this figure has been built with a Matlab program).
$A(0, 0, 1)$, $B(0, 1, 0)$ 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 $\sigma=\dfrac{\pi}{4}+\dfrac{\pi}{2}+\dfrac{\pi}{2}-\pi=\dfrac{\pi}{4}$.