Question 1054093
<pre><b>

{{{drawing(400,400,-2,6,-2,6,
line(0,0,3sqrt(2)/2,3sqrt(2)/2),line(0,0,5,0),
line(5,0,-8,13),arc(0,0,6,-6,10,100),
line(0,-.1,0,.1),line(1,-.1,1,.1),line(2,-.1,2,.1),line(3,-.1,3,.1),line(4,-.1,4,.1),line(5,-.1,5,.1), arc(0,0,2,-2,40,50), arc(0,0,4,-4,43,47),
red(arc(5,0,2.6,-2.6,135,180),locate(4.13,.36,"45°")), locate(0,0,A),
locate(5,0,B),locate(2.13,2.43,"C?")




 )}}}

In the above drawing, each of the tic-marks are equal and
represent 1 inch each.  The angle B has measure 45°.  We can
see by the arc that the line AC, which equals 3 inches, is 
not long enough to reach the slanted side of the 45° angle.
Therefore triangle ABC is not possible.  We can also show
by the law of sines that no triangle ABC with the given
properties in possible.

{{{b/sin(B)=c/sin(C)}}}

{{{b*sin(C)=c*sin(B)}}}

{{{sin(C)=(c*sin(B))/b}}}

{{{sin(C)=(5*sin("45°"))/3}}}

{{{sin(C)="1.178511302"}}}

Since sines of angles are always less than one, this shows
that there is no possible way to have an angle C.  Therefore
it is impossible to have a triangle ABC with the given
properties.

Edwin</pre><b>