Is there a similar formula in spherical and hyperbolic geometry as Euclidean Geometry? 
*

*In an Euclidean plane,  we know that the area of a triangle is determined by the length of base and the height, then is there a similar thing do happen in Spherical and hyperbolic spaces?

*In Euclidean plane, we know that the cosine law says that (suppose $\alpha$, $\beta$, $\gamma$ are the angles, and $a, b, c$ are the lengths opposite $a, b, c$ respectively.)
$$
\cos\gamma=\frac{a^2+b^2-c^2}{2ab}.
$$
Then is there a analogue in spherical and hyperbolic geometry? I have noted that the First and Second Cosine law are not so clearly relevant with this formula.
 A: Yes,there are analogues in spherical and hyperbolic geometry.
Start with a right angled triangle $\gamma=\pi/2$.
$$\cos(c/R)=cos(a/R) cos(b/R) $$
can be expanded in power series:
$$( 1-(c/R)^2/2)\approx  (1-(a/R)^2/2)*(1-(b/R)^2/2)  $$
and let R go to infinity to derive Pythagoras theorem approaching from spherical side getting
$ c^2 = a^2 + b^2 $
Likewise,
$$\cosh(c/R)=\cosh(a/R) \cosh(b/R) $$
expand cosh, let R go to infinity to derive it approaching from the hyperbolic side:
$ c^2 = a^2 + b^2 $
We further note that the $R`s$ are radii of the Sphere and Pseudosphere respectively. Pythagoras theorem holds in the Flat plane.
Cosine Law Spherical Trig
Cosine Law Hyperbolic Trig
A unified or "pan-geometry" cosine formula for both  unit spherical or pseudospherical radius can be written:
$$ \cos(h) c= \cos(h)a \cos(h)b \pm \sin(h)a \sin(h)b \cos C. $$
A: The amount you turn in radians when going around any shape is precisely 2$\pi$ minus the product of the area of the shape and the amount of curvature. The amount you turn when going around a triangle is the sum of the exterior angles. The sum of the interior angles in radians therefore is $\pi$ plus the product of the area and the curvature.
