Is it possible to split a triangle in two, making a two right triangles, and how? I have a problem, where I want to figure out how to split a triangle in two making two right triangles. The line has to be 90 degrees from line c(red font in picture), and has to meet with point C(black font)
We know that The triangle is 73degrees, 75 degrees, and 32 degrees, we also know that line c is 8cm. Is it possible to split it through mathematics?

Need help with this
The triangles are well defined.  For each you know all the angles (75,90,15) and (72,90,18). You can use the law of sines to get the side lengths when you know one side.
I saw nobody answered this and the few "starter" answers were wrong.

I began explaining a solution where the first step is to divide the triangle into a small right triangle using the left side (T1 in the diagram) to define the height of the triangle using angles of 73, 90 and 17 (all must equal 180 and thus we know the acute angle at the top) and solving for the base of triangle T1 using sin(17) = x/8.  (sine t = opposite/base)

From here, using the hypotenuse and base, the leg, or height is calculated as sqrt(8^2 - 2.34^) = 7.65

Now that we also have the base leg of T2 we can solve for the hypotenuse using the acute angle 32 degrees as sin(32) = 7.65/x - 14.44 and solving for the missing side of T2.

Next add the missing side of T2 to the base of T1 and the result is 2 known hypotenuse lengths of 2 new triangles NT1 (73, 90, 17) and NT2 (75, 90, 15) which are right triangles meeting on line C. 

Use each acute angle to find the base as above and you end up with base of NT1 of 4.26 and a base leg of NT2 of 3.74 mathematically bisecting the 8cm side at two right triangles.

*** NOTE - I saw on the image of the solution there was a line which states
   "further by dividing both sides by cos(32)" 
which should state 
   "further by dividing both sides by sin(32)"

I corrected that, labelled and color coded the labels for the 2 subsets of triangles and side lengths and added a Pythagorean theorem for sub triangle T2 as proof that law of sines provided the correct answer.

I saw a down vote and took the optimistic approach that the person couldn't follow the solution due to the mentioned typo and lack of labels rather than could not follow geometry and trigonometry.

one solution