A triangle has sides A, B, and C. Sides A and B are of lengths #12# and #5#, respectively, and the angle between A and B is #pi/12#. What is the length of side C?

We use Cosine-rule for the given triangle & get

#C^2#=#A^2+B^2-2AB(cosangle betwn. A & B)#
#=12^2+5^2-2*12*5*cos(pi/12)#
#=144+25-120{sqrt(1/2(1+cos(2*pi/12))}#
#=169-120{sqrt(1/2(1+sqrt3/2))}#
#=169-120{sqrt{(2+sqrt3)/4)}#
#=169-120/2{sqrt((2+sqrt3))}#
#=169-60(sqrt(2+sqrt3))#
#=169-60(sqrt((4+2sqrt3)/2)#
#=169-(60/sqrt2)*{sqrt(3+1+2sqrt3)}#
#=169-(60*sqrt2)/2*[sqrt{(sqrt3)^2+(sqrt1)^2+2*sqrt3*sqrt1}]#
#=169-30*sqrt2*sqrt[(sqrt3+sqrt1)^2]#
#=169-30*sqrt2*(sqrt3+1)#

Always a good start to draw a diagram or sketch.

#color(blue)("Method Plan")#

Objective: determine #/_b# then by using the #sin(/_b) # determine C
'............................................

Step 1: #-># determine h; then cP using #sin(/_c)#

Step 2:#-># 12 - cP = bP

Step 3:# -> /_b = arctan(h/(bP))#

Step 4: #=> sin(/_b) = h/C " "=>" " C=h/sin(/_b)#

To reduce accumulated rounding error it is better do the calculation only at the end.
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Answering the question")#

Step 1: #" "color(brown)(h=Bsin(/_c)->5sin(pi/12))#

#" "color(brown)(cP = Bcos(/_c) -> 5cos(pi/12))#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Step 2:#" "color(brown)(bP=12-5cos(pi/12))#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Step 3: #" "color(brown)( /_b = arctan(h/(bP))->/_b=arctan((5sin(pi/12))/(12-5cos(pi/12))))#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Step 4: #" "color(brown)(C=h/sin(/_b)->color(blue)(C=h/[sin[arctan((5 sin(pi/12))/(12-5cos(pi/12))) ]))#

#color(green)("You will see arctan on your calculator as "tan^(-1))#
Arctan returns angle that made that made the tangent value

So if #tan(theta)=x" "# then #" "arctan(x)=theta#