A horizontal line passing through point A is L1.
A horizontal line passing through point B & C is L2.

Given Triangle is Inverted :
                     __B__________C___ L2 
                        \________/
			 \______/              Fig: An Outline Of inverted Triangle
			  \____/
                           \__/
                      ______\/_______L1      
                             A

          Look for triangles with base parallel to L1(White Colored).You will find:
          First level with height 1 Unit: 0
                  Second Level with height 1 units: 1 
                  In general for nth level with height 1 unit: n-1
                  So total number of triangles with height of 1 unit are: 
                                   0+1+2+3..................n-1 = n(n-1)/2
          First level with height 2 unit: 0
                  Second level with height 2 unit: 0
                  Third level with height 2 unit: 0
                  Fourth level with height 2 unit: 1
                  Fifth level with height 2 unit : 2
                  In general nth level with height 2 units is: n-3
                  So number of triangles with height 2 units are: 1+2+3+.............n-3= 
                 (n-3)(n-2)/2

In general Total number of triangles with the given triangle inverted & with base parallel to L1 are:
                  1/2*{ n(n-1)+(n-2)(n-3)+(n-4)(n-5)+.............................0}   -------Equation1
 

Original Triangle:
                 Look for triangles with base parallel to L2 from top:

                  Number of triangles with  height 1(filled with white color) at first to nth level:  
                                          1+2+3+----------------+n= n(n+1)/2
                  Number of triangles with height 2 from first to nth level:
                                           0+1+2+3+----------------+(n-1)=(n-1)(n)/2
                  Number of triangles with height 3 from first to nth level:
                                           0+0+1+2+3+--------------+(n-2)=(n-2)(n-1)/2

In general total number of triangles with base parallel to L2 are:
                  1/2*{ n(n+1)+(n-2)(n-3)+(n-4)(n-5)+...........................0}------------Equation2

So total number of triangles in given triangle seen from every angle are(Equation1 + Equation2:

  1/2*[{n(n-1)+(n-2)(n-3)+(n-4)(n-5)+............0}+{ n(n+1)+(n)(n-1)+(n-2)(n-1)+............0}]

  Now the task is to simply the above equation

  =1/2* {n(n+1)+(n-2)(n-1)+..................}+{(n)(n-1)+(n-2)(n-3)+..................

  =1/2*{(n)(3n-1)+(n-2)(3n-7)+(n-4)(3n-13)+...........}

  =1/2*S(3n^2-12nr-n+12r^2+2r)     where r =0 to X & X=int(n+1)/2

  =1/2{3*n^2*X - 12n.X(X+1)/2-nX+12* X(X+1)(2X+1)/2+2*X(X+1)/2}

On simplification,
   
    Total Number of Triangles : X/2*{(4X^2-5X+1)+n(3n+5-6X)} where 
                       n= Number of levels
                       X=integer value of (n+1)/2
    
Example:
         Lets take n=7 then X=(7+1)/2=4
         Total Number of triangles: 4/2*{(4*4^2-4*5+1)+7(3*7+5-6*4)}
				   =2*{45+14}
				   =118
         For n=1      N= 1
             n=2      N= 5
             n=3      N= 13
             n=4      N= 27
             n=5      N= 48
             n=6      N= 78