geometry inequality triangles help if you can
posted by Knights .
Two sides of an obtuse triangle are 16 and 21. How many possible lengths are there for the third side, if it is a positive integer?
i know that in an obtuse triangle
a^2+b^2<c^2
a+b>c
but i tried plugging in and it wont work so couldyou guys help me?
btw merry christmas and happy new year

First let's establish the domain for the third side
let the third side be c
c+16>21 > c > 5
c+21 > 16 > c > 5 , obviously
21 + 16 > c > c < 37
so to be a triangle in the third side must be
5 < c < 37
Now to be obtuse
c^2> 21^2+16^2
c^2 > 697
c > 26
could one of the other angles be obtuse ?
21^2 >c^2 + 16^2
c^2 < 185
c < 14
16^ > c^2 + 21^2
c^2 < a negative, which is not possible
So for the angle opposite the third side to be obtuse,
26 < c < 37
For the angle opposite the side 21 to be obtuse,
5 < c < 14 
thanks a lot the answer is 18 right?
merry christmas and happy new year to y 
No, 18 does not work.
Notice my two possible solutions, .....
So for the angle opposite the third side to be obtuse,
26 < c < 37
So the third side could be 27, 28, 29, 30 31 32 33 34 35 or 36
For the angle opposite the side 21 to be obtuse,
5 < c < 14
so the third side could be 6 7 8 9 10 11 12 or 13
make a sketch using one of the answers and illustrate that the angle must be obtuse by using the cosine law.
Also test it with your 18 to show it does not work 
I meant the total number of possibilities. Thanks anyway!

Eighteen is correct. I had the same problem and got it right