An angle bisector of a triangle divides the opposite side of the triangle into segments 6 cm and 5cm long. a second side of the triangle is 6.9 cm long. find the longest and shortest possible lengths of the third side of the triangle. round answers to the nearest tenth of a centimeter.

There is a not so well-known theorem that solves this problem.

The theorem is stated as follows:
"Each angle bisector of a triangle divides the opposite side into segments proportional in length to the adjacent sides"  (Coxeter & Greitzer)

This means that for a triangle ABC, where angle A has a bisector AD such that D is on the side BC, then 
BD/DC=AB/AC

Here either
BD/DC=6/5=AB/AC, where  AB=6.9,  
then we solve for  AC=AB*5/6=5.75,

or

BD/DC=6/5=AB/AC, where  AC=6.9,  
then we solve for AB=AC*6/5=8.28

Hence, the longest and shortest possible lengths of the third side are
8.28 and 5.75 units respectively.

The answer is budget i believe. hope this but it’s self explanatory.