Question 696752
<pre>
Let's draw it:

{{{drawing(400,400/3,-1,11,-1,3, triangle(0,0,3cos(.7227342478),3sin(.7227342478),10,0),

circle(cos(.7227342478), sin(.7227342478),.1),
circle(2cos(.7227342478),2sin(.7227342478),.1),
circle(3cos(.7227342478),3sin(.7227342478),.1),
circle(0,0,.1),

circle(10-cos(.2506556623),sin(.2506556623),.1),
circle(10-2cos(.2506556623),2sin(.2506556623),.1),
circle(10-3cos(.2506556623),3sin(.2506556623),.1),
circle(10-4cos(.2506556623),4sin(.2506556623),.1),
circle(10-5cos(.2506556623),5sin(.2506556623),.1),
circle(10-6cos(.2506556623),6sin(.2506556623),.1),
circle(10-7cos(.2506556623),7sin(.2506556623),.1),
circle(10-8cos(.2506556623),8sin(.2506556623),.1),

circle(1,0,.1),
circle(2,0,.1),
circle(3,0,.1),
circle(4,0,.1),
circle(5,0,.1),
circle(6,0,.1),
circle(7,0,.1),
circle(8,0,.1),
circle(9,0,.1),
circle(10,0,.1),
locate(-.3,0,B),locate(10.2,0,C), locate(2.25,2.6,A)

)}}}

Looks like &#8736;A is an obtuse angle, making &#5123;ABC an obtuse triangle.

But your geometry teacher will not accept "draw it, look at it 
and see", even though that is more convincing than the Pythagorean
theorem!  lol

So we test it to see if &#5123;ABC is a right triangle.  If
it is then the LONGEST side BC, will be the hypotenuse,
and &#8736;A, across from it, will be a right angle.

So we calculate the hypotenuse of a right triangle whose legs
are 3 and 8, and see if we get 10.

c² = a² + b²
c² = 3² + 8²
c² = 9 + 64
c² = 73
c = &#8730;<span style="text-decoration: overline">73</span>
c = 8.544, approximately.

So &#5123;ABC is not a right triangle, and since 10 is greater than
8.544, that means that &#8736;A has to open wider than 90°,
which makes it an obtuse angle, and therefore &#5123;ABC is obtuse.

[Note: If it had turned out that c had been greater than the
longest side, the angle would not have opened as wide as 90°,
and it would have been acute. And if it had turned out to be
the same, it would have been a right angle. Neither of those
were the case here, but in other problems they might be.]

Edwin</pre>