A triangle has sides A, B, and C. Sides A and B have lengths of 8 and 12, respectively. The angle between A and C is #(3pi)/4# and the angle between B and C is # pi/3#. What is the area of the triangle?

No no no! Triangles have sides #a,b,c# and vertices #A,B,C#. Here # a=8, b=12,# #B={3\pi}/4,# #A=\pi/3#.

Yet another trig problem based only on angles related to 30 and 45 degrees.

This can't be a real triangle because:

# a/sin A ne b / sin B#

#a/sin A = 8/sin(pi/3) = 8/(\sqrt{3}/2) = 16/sqrt{3}#

#b/sin A= 12/sin(3pi/4) = 12/(sqrt{2}/2) = 24/sqrt{2} #

They're not equal.

That's too bad, because the area would have been calculated as:

# text{area} = 1/2 a b sin C #

# = 1 /2 (8)(12) sin(pi - {3pi}/4 - pi/3) #

# = 48 sin( {3 pi}/4 - pi/3) #

# = 48 ( sin({3pi}/4) cos(pi/3) - cos({3pi}/4) sin(pi/3) ) #

#= 48 ( (\sqrt{2}/2) (1/2) - (- \sqrt{2}/2) (\sqrt{3}/2) ) #

# = 12(sqrt{2} + sqrt{6}) #

But it's not.

#a = 8, hat A = pi/3, b = 12, hat B = (3pi)/4#

#color(maroon)("THEOREM. A greater angle of a triangle is opposite a greater side. "#
#color(brown)("Let ABC be a triangle in which angle ABC is "#

#color(brown)("greater than angle BCA; then side AC is also greater than side AB"#

#color(crimson)("But though " a < b, hat A > hat B#

#color(crimson)("Such a triangle cannot exist"#