Question 1037821
∆ABC with m∠A = 38° and AB = 6.3 and BC = 4.2
<pre><b>
Using law of sines:

{{{drawing(400,300,-1,8,-1,5,
red(arc(0,0,2.3,-2.3,0,38)),locate(3.35,0,"C'"),locate(6.57,0,C),
locate(5,4.2,B),locate(0,0,A), locate(5.85,2,4.2),
triangle(0,0,6.3cos(38*pi/180),6.3sin(38*pi/180),3.35328,0),
locate(.43,.35,"38°"),locate(1.7,2,6.3),locate(3.5,2,4.2),
triangle(0,0,6.3cos(38*pi/180),6.3sin(38*pi/180),6.57566,0) )}}}

{{{matrix(1,3,BC,or,"BC'")/sin("<A")}}}{{{""=""}}}{{{AB/sin(matrix(1,3,"<C",or,"<AC'B"))}}}

{{{4.2/sin("38°")}}}{{{""=""}}}{{{6.3/sin(matrix(1,3,"<C",or,"<AC'B"))}}}

{{{4.2sin(matrix(1,3,"<C",or,"<AC'B"))}}}{{{""=""}}}{{{6.3sin("38°")}}}

{{{sin(matrix(1,3,"<C",or,"<AC'B"))}}}{{{""=""}}}{{{6.3sin("38°")/4.2}}}

{{{sin(matrix(1,3,"<C",or,"<AC'B"))}}}{{{""=""}}}{{{0.923492213}}}

&#8736;C is an acute (QI) angle and &#8736;AC'B is an 
obtuse (QII) angle.  They have the same sine and are
supplementary.

Using the inverse sine feature of our calculator,

&#8736;C = 67.44208077°

so &#8736;AC'B = 180°-67.44208077° = 112.5579192°

To find side AC and AC', we use the law of cosines:

{{{CB^2}}}{{{""=""}}}{{{AC^2+AB^2-2*AB*AC*cos("<A")}}}

Swap sides:

{{{AC^2+AB^2-2*AB*AC*cos("<A")}}}{{{""=""}}}{{{CB^2}}}

Rearrange terms and a factor with 0 on the right:

{{{AC^2-2*AB*cos("<A")*AC+AB^2-CB^2}}}{{{""=""}}}{{{0}}}

{{{AC^2-2*6.3*cos("<A")*AC+6.3^2-4.2^2}}}{{{""=""}}}{{{0}}}

{{{AC^2-12.6*cos("38°")*AC+22.05}}}{{{""=""}}}{{{0}}}

Use the quadratic formula

{{{AC}}}{{{""=""}}}{{{(-(-12.6*cos("38°")) +- sqrt( (-12.6*cos("38°"))^2-4*1*2.05 ))/(2*1) }}}

{{{AC}}}{{{""=""}}}{{{((12.6*cos("38°")) +- sqrt( (-12.6*cos("38°"))^2-4*2.05 ))/2 }}}

Using the + gives us AC = 6.575659859
Using the - gives us AC' = 3.353275637

We can find &#8736;ABC and &#8736;ABC' by adding the 
two angles we now know and subtracting from 180°.

You can finish that part.

Edwin</pre></b>