Answer:
It's obtuse.
Step-by-step explanation:
Mumble. Smells al-kashi laws of cosine to me.
[tex]a^2-b^2-c^2 = -2bc cos\alpha\\b^2-c^2-a^2=-2ca cos \beta\\c^2-a^2-b^2=-ab cos \gamma[/tex]
Now, we don't need to calculate the RHSs, and then:
If at least one of the three LHSs is greater than zero, the triangle is obtuse
If at least one is zero, it's a right triangle (pythagorean theorem anyone?)
if all three numbers are less than zero, the triangle is acute
Start crunching numbers.
[tex]139-73-36 = 30[/tex] First try was the charm. finding 30 means that the product [tex][-2(\sqrt{73})(6) ]cos \alpha[/tex] is positive, and since the product in the large bracket is negative, the cosine has to be 0, thus the angle has to be greater than 90°, QED
