how do you solve a trianglular arithmagon with the numbers give 38, 48, and 68

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In the triangle ABC, side AB=38, side AC=48, side BC=68.

If 38, 48 and 68 are the numbers on the sides we add them together: 38+48+68=154.   

Now divide by 2: 154/2=77.

The vertex C (corner) opposite the side AB=38 will be 77-38=39.

The vertex B opposite the side AC=48 will be 77-48=29.

The vertex A opposite the side BC=68 will be 77-68=9.

         9

    38  48

29   68   39
by Top Rated User (1.2m points)

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