The number of acute angled triangles whose sides are three consecutive positive integers and whose perimeter is at most 100 is

Since the sides are three consecutive integers, let the sides be n, n+1, and n+2, where n > 0.

The sum of the two sides is greater than the third side.

=> $$n+n+1>n+2$$

=> $$n>1\rightarrow1$$

The perimeter of the triangle is at most 100.

=> $$n+n+1+n+2\le100$$

=> $$3n+3\le100$$

=> $$n\le32.33\rightarrow2$$

Also, the given triangle is an acute angled triangle.

=> $$\left(n+2\right)^2< n^2+\left(n+1\right)^2$$

=> $$n^2+4n+4< n^2+n^2+2n+1$$

=> $$n^2-2n-3>0$$

=> $$\left(n-3\right)\left(n+1\right)>0$$

=> $$n<-1\ or\ n>3\rightarrow3$$

Thus, comparing eq. 1, 2 and 3, we will get -

=> $$3<n\le32.33$$

The values of n are - $$n=4,5,6,7,......32$$. Thus, there are 29 values possible.