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^@ #