# Math Help - obtuse triangle help

1. ## obtuse triangle help

I have an obtuse triangle. It has sides 6, 8, and x. The angle opposite side x is an obtuse angle. What could be some possible values of side x?

I know there is a rule for obtuse angles that is: (a^2) + (b^2) < c^2.
So could 15 be an answer? 36 + 100 < 225

Thanks for helping!

2. Originally Posted by curio03
I have an obtuse triangle. It has sides 6, 8, and x. The angle opposite side x is an obtuse angle. What could be some possible values of side x?

I know there is a rule for obtuse angles that is: (a^2) + (b^2) < c^2.
So could 15 be an answer? 36 + 100 < 225

Thanks for helping!
if c is the side opposite the obtuse angle ...

$\sqrt{6^2 + 8^2} < c < 14$

now ... where did the value "14" come from?

3. That rule is for right-triangles. So the lower bound will be 10. Can you use the triangle inequality to deduce the upper bound ?

4. Originally Posted by curio03
So could 15 be an answer?
Try constructing a triangle with sides 6-8-15

5. Originally Posted by Traveller
That rule is for right-triangles. So the lower bound will be 10. Can you use the triangle inequality to deduce the upper bound ?
No, there is a "<" in the formula, not "=". The rule given, $a^2+ b^2< c^2$ is correct for obtuse angles opposite side c. I have no idea where you got "10" as a lower bound. The lower bound derived from $a^2+ b^2< c^2$ is $\sqrt{6^2+ 8^2}$ as skeeter said. And that is slightly less than 10.

6. Originally Posted by HallsofIvy
No, there is a "<" in the formula, not "=". The rule given, $a^2+ b^2< c^2$ is correct for obtuse angles opposite side c. I have no idea where you got "10" as a lower bound. The lower bound derived from $a^2+ b^2< c^2$ is $\sqrt{6^2+ 8^2}$ as skeeter said. And that is slightly less than 10.
True, I didn't notice the inequality.

An obtuse angle may be larger than a right angle by an arbitrarily small amount. Hence the greatest lower bound is 10, as in that side will definitely exceed 10 and might be smaller than any value greater than 10. I should have explained this clearly.