<PRE><B>Question 862732</B>
Check first for a fit to Pythagorean Theorem relationship.

{{{37^2}}} versus {{{12^2+34^2}}}
1369 versus 1300 are not equal.


Maybe not the best way, but workable:
{{{a^2+b^2-2ab*cos(s)=c^2}}}
{{{-2ab*cos(s)=c^2-a^2-b^2}}}
{{{highlight_green(cos(s)=(a^2+b^2-c^2)/(2ab))}}}
Is the angle opposite the 37 greater than 90 degrees?  c=37.
{{{cos(s)=(12^2+34^2-37^2)/(2*12*34)}}}
{{{cos(s)=(-69)/(816)}}}
Obviously a small negative number, so s is in quadrant number 2, meaning s is like {{{90&lt;s&lt;180}}} using degrees measure.  Angle s is obtuse.


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ANSWER:   OBTUSE TRIANGLE.
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