Here is a reference for The Ambiguous Case
#angle A# is acute. Compute value of h:
#h = (c)sin(A)#
#h = (10)sin(60^@)#
#h ~~ 8.66 #
#h < a < c#, therefore, two possible triangles exist, one triangle has #angle C_("acute")# and the other triangle has #angle C_("obtuse")#
Use The Law of Sines to compute #angle C_("acute")#
#sin(C_("acute"))/c = sin(A)/a#
#sin(C_("acute")) = sin(A)c/a#
#C_("acute") = sin^-1(sin(A)c/a)#
#C_("acute") = sin^-1(sin(60^@)10/9)#
#C_("acute") ~~ 74.2^@#
Find the measure for angle B by subtracting the other angles from #180^@#:
#angle B = 180^@ - 60^@ - 74.2^@#
#angle B = 45.8^@#
Use the Law of Sines to compute the length of side b:
side #b = asin(B)/sin(A)#
#b = 9sin(45.8^@)/sin(60^@)#
#b ~~ 7.45#
For the first triangle:
#a = 9,b ~~ 7.45, c = 10, A = 60^@, B ~~ 45.8^@, and C ~~ 74.2^@ #
Onward to the second triangle:
#angle C_("obtuse") ~~ 180^@ - C_("acute")#
#C_("obtuse") ~~ 180^@ - 74.2^@ ~~ 105.8^@#
Find the measure for angle B by subtracting the other angles from #180^@#:
#angle B = 180^@ - 60^@ - 105.8^@ ~~ 14.2^@#
Use the Law of Sines to compute the length of side b:
#b = 9sin(14.2^@)/sin(60^@)#
#b ~~ 2.55#
For the second triangle:
#a = 9,b ~~ 2.55, c = 10, A = 60^@, B ~~ 14.2^@, and C ~~ 105.8^@ #
