Question 819293
{{{drawing(400,200,-2,18,-2,8,
triangle(0,0,10.39,0,10.39,6),
rectangle(10.39,0,10.09,0.3),
triangle(-3,0,-3,30,30,0),
locate(-0.1,-0.1,B),locate(10.3,6.7,C),
locate(10.3,-0.1,D),locate(1.1,1.1,30^o),
red(arc(0,0,5.5,5.5,330,360)),
locate(10.5,3.5,6),locate(5,2.88,12),
blue(line(15.68,0,10.39,6)),locate(12.7,3,blue(8)),
green(line(5.1,0,10.39,6)),locate(7.7,3,green(8)),
locate(15.58,0,blue(Y)),locate(5,0,green(X))
)}}} ABC could be XBC OR YBC. vertex A could be X or Y, then side b (AB) would be XC or YC, measuring 8cm.
Since there were two possible triangles, I drew a third triangle, right triangle BCD.
CD = BC{{{sin(30^o)=12*0.5=6}}}
We can also calculate
BD = BC{{{cos(30^o)=12*(sqrt(3)/2)=6sqrt(3)=about10.39}}}
We can calculate the length of XD = YD in congruent right triangles XDC and YDC.
That length is
XD = YD ={{{sqrt(8^2-6^2)sqrt(64-36)=sqrt(28)=2sqrt(7)=about5.29}}}
Then,
BX = BD - XD ={{{highlight(6sqrt(3)-2sqrt(7))}}}= about{{{highlight(5.10)}}}
BY = BD + YD ={{{highlight(6sqrt(3)+2sqrt(7))}}}= about{{{highlight(15.68)}}}
So the approximate measure of the third side is
either {{{highlight(5.10cm)}}} or {{{highlight(15.68cm)}}} .
 
ALTERNATE SOLUTION:
Maybe your teacher expected you to use law of cosines,
{{{b^2=a^2+c^2-2ac*cos(B)}}}
{{{8^2=12^2+c^2-2*12*c*cos(30^o)}}}
{{{64=144+c^2-2*12*c*(sqrt(3)/2)}}}
{{{64=144+c^2-(12sqrt(3))c}}}
{{{c^2-(12sqrt(3))c+144-64=0}}}
{{{c^2-(12sqrt(3))c+80=0}}}
That quadratic equation can be solved using the quadratic formula:
{{{c = (12sqrt(3) +- sqrt((12sqrt(3))^2-4*1*80 ))/(2*1)=(12sqrt(3) +- sqrt(12^2*3-320 ))/2=(12sqrt(3) +- sqrt(432-320))/2=(12sqrt(3) +- sqrt(112))/2=(12sqrt(3) +- sqrt(16*7))/2=(12sqrt(3) +- 4sqrt(7))/2=highlight(6sqrt(3) +- 2sqrt(7))}}}
The quadratic equation can also be solved by completing the square:
{{{c^2-(12sqrt(3))c+80=0}}}
{{{c^2-(12sqrt(3))c=-80}}}
{{{c^2-(12sqrt(3))c+(6sqrt(3))^2=(6sqrt(3))^2-80}}}
{{{(c-6sqrt(3))^2=6^2*3-80}}}
{{{(c-6sqrt(3))^2=108-80}}}
{{{(c-6sqrt(3))^2=28}}} so {{{system(c-6sqrt(3)=sqrt(28)=2sqrt(7), "OR",c-6sqrt(3)=sqrt(28)=-2sqrt(7))}}} ,
leading to the solutions {{{highlight(c=6sqrt(3) +- 2sqrt(7))}}}
 
ANOTHER ALTERNATE:
Since you have the measures of angle B and side b, you can apply law of sines, and find {{{sin(A)}}}, and two possible approximate measures for angle A.
Then you could calculate the approximate measures for the two options for angle C, and for {{{sin(C)}}}, and then use law of sines again to find the two possible measures for side c.