obtuse triangle

[iullyan]

"Prove that the triangle with sides 3,5,6 is obtuse". I've tried with the law of cosines to find out some angles and then prove that an angle is bigger than 90 degrees but it resulted some reports which didn't help me too much. Can you give me an idea?

 

[stapel]

"Prove that the triangle with sides 3,5,6 is obtuse". I've tried with the law of cosines to find out some angles and then prove that an angle is bigger than 90 degrees but it resulted some reports which didn't help me too much. Can you give me an idea?

Obviously, the angle opposite the side with the longest length will have the largest measure. Please reply showing what you did and the results you obtained.

Please be complete. Thank you!

 

[Ishuda]

"Prove that the triangle with sides 3,5,6 is obtuse". I've tried with the law of cosines to find out some angles and then prove that an angle is bigger than 90 degrees but it resulted some reports which didn't help me too much. Can you give me an idea?

The law of cosines says
c² = a² + b² - 2 a b cos(C)
where C is the angle opposite the side c. So what did you get for the three angles (or the cosines of the three angles)?

 

[iullyan]

The law of cosines says
c² = a² + b² - 2 a b cos(C)
where C is the angle opposite the side c. So what did you get for the three angles (or the cosines of the three angles)?

For cosA I obtained -1/15, for cosB=5/9 and for cosC=13/15.

 

[Ishuda]

For cosA I obtained -1/15, for cosB=5/9 and for cosC=13/15.

Now take the arc-cosine of each of those values if you want the angles. However you can also note that the cosine is non-negative between 0 and \(\displaystyle \frac{\pi}{2}\) and non-positive between \(\displaystyle \frac{\pi}{2}\) and \(\displaystyle \pi\). Thus the angle with the negative value for the cosine is greater than \(\displaystyle \frac{\pi}{2}\) (90 degrees).

 

[iullyan]

Now take the arc-cosine of each of those values if you want the angles. However you can also note that the cosine is non-negative between 0 and \(\displaystyle \frac{\pi}{2}\) and non-positive between \(\displaystyle \frac{\pi}{2}\) and \(\displaystyle \pi\). Thus the angle with the negative value for the cosine is greater than \(\displaystyle \frac{\pi}{2}\) (90 degrees).

Thank you.