Note: I solved for points randomly "in" the sphere not "inside" the sphere. I red it too fast!

Let be (a, b, c) = three natural numbers standing for angles of a generic triangle, so that:

a b c always > 0

and 

the values of a b c are not repeated in different order. (F ex.we do not admit both: (35,69,76) and (69,35,76) but just one df them). 

Then (a,b,c) perform all the possible "forms" of natural triangles. We can count manually the total number of triangles for all the values of a, b, c. [Obviously a+b+c =180°]

They are 2700. 

 

675 of them verify that a, b, c are all < 90. So these are rectangles triangles or acute triangles (632 acute & 43 rectangles). 

Then the % of acute is 632/2700 = 23,3%

But on the puzzle we wont have natural angles but real angles. The probability of a rectangle triangle on the sphere is infinitesimal (lim ->0) and all 675 natural

 forms will become on the sphere acute triangles).

My conclusion is then that the probability of finding random acute triangles on the sphere is

675/2700 = 0.25