Question 1207648
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Refer to the diagram tutor Edwin has drawn.
{{{drawing(405,1568/9,-40,185,-10,88,

triangle(-30.82762166,66.110048,120,0,0,0),
triangle(0,0,120,0,174.8922588,78.39426942),
triangle(-30.82762166,66.110048,174.8922588,78.39426942,174.8922588,78.39426942),

locate(0,0,C), locate(120,0,D),
locate(-31,75,A),locate(175,87,B), locate(50,0,matrix(1,2,120,ft)),

locate(57,36,E),


red(arc(0,0,50,-50,0,115),locate(7,35,115^o)),blue(arc(0,0,39,-39,23,115),locate(-2.7,17,92^o)),

red(arc(120,0,50,-50,55,180),locate(110,38,125^o)),blue(arc(120,0,39,-39,55,155),locate(108,19,100^o)),

green(arc(0,0,60,-60,0,23),locate(30,16,23^o),
arc(120,0,60,-60,155,180),locate(77,13,25^o)

)

)}}}


There are a lot of triangles to keep track of, so it might be helpful to peel the triangles ACD and BCD apart to get this
{{{drawing(605,1568/9,-40,345,-10,88,

triangle(-30.82762166,66.110048,120,0,0,0),
triangle(0+150,0,120+150,0,174.8922588+150,78.39426942),


locate(0,0,C),locate(0+150,0,C), locate(120,0,D),locate(120+150,0,D),
locate(-31,75,A),locate(175+150,87,B), locate(50,0,matrix(1,2,120,ft)),locate(50+150,0,matrix(1,2,120,ft)),


locate(7,35-20,115^o),

locate(110+140,38-20,125^o),

locate(30+150,16,23^o),
locate(77,13,25^o)

)}}}
The diagrams aren't to scale.


Recall that for any triangle, the interior angles always add to 180 degrees.
We'll use this fact to find the missing angles of triangles ACD and BCD.


For triangle ACD, the missing angle is A = 180-C-D = 180-115-25 = 40 degrees.
For triangle BCD, the missing angle is B = 180-C-D = 180-23-125 = 32 degrees.


Let's update the diagram
{{{drawing(605,1568/9,-40,345,-10,88,

triangle(-30.82762166,66.110048,120,0,0,0),
triangle(0+150,0,120+150,0,174.8922588+150,78.39426942),


locate(0,0,C),locate(0+150,0,C), locate(120,0,D),locate(120+150,0,D),
locate(-31,75,A),locate(175+150,87,B), locate(50,0,matrix(1,2,120,ft)),locate(50+150,0,matrix(1,2,120,ft)),


locate(7,35-20,115^o),

locate(110+140,38-20,125^o),

locate(30+150,16,23^o),
locate(77,13,25^o),


locate(7-25,35+20,40^o),
locate(7+280,35+30,32^o)

)}}}


From here, we have a few pathways we could take. 
The path I'll take is to determine the side AC (from triangle ACD) and determine side BC (from triangle BCD). 
Use the Law of Sines to find these lengths.


I'll get you started with the setup equations
AC/sin(D) = CD/sin(A)
AC/sin(25) = 120/sin(40)
and
BC/sin(D) = CD/sin(B)
BC/sin(125) = 120/sin(32)
Make sure that your calculator is set to degrees mode.

I'll let the student finish this subsection.


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Let's focus on triangle ABC.
{{{drawing(405,1568/9,-40,185,-10,88,
triangle(-30.82762166,66.110048,174.8922588,78.39426942,0,0),

locate(0,0,C),
locate(-31,75,A),locate(175,87,B),

locate(-2.7,17,92^o)

)

)}}}


The last batch of steps would be to use the Law of Cosines to find the length of side AB.
x = length of AC
y = length of BC
z = length of AB


Law of Cosines
z^2 = x^2 + y^2 - 2*x*y*cos(C)
z^2 = x^2 + y^2 - 2*x*y*cos(92)


I'll let the student finish up.
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