Question 697680
<pre>
Suppose you know one side and two angles, say,
AB = 10, Angle B = 42°, Angle C = 32° 

{{{drawing(700,300,-1,3,-1,2,
locate(.3,1,10),
locate(1,1.9,A), locate(0,0,B),locate(2.8,0,C),
locate(.2,.25,"42°"),locate(2.2,.25,"32°"),
triangle(0,0,1+sqrt(3),0,1,sqrt(3))   )}}}

Draw perpendicular AD from A to BC

{{{drawing(700,300,-1,3,-1,2, locate(.95,0,D),
locate(.3,1,10), green(line(1,sqrt(3),1,0)),
locate(1,1.9,A), locate(0,0,B),locate(2.8,0,C),
locate(.2,.25,"42°"),locate(2.2,.25,"32°"),
triangle(0,0,1+sqrt(3),0,1,sqrt(3))   )}}}

{{{AD/AB}}} = sin(42°)
{{{AD/10}}} = sin(42°)
AD = 10·sin(42°)
AD = 6.69

{{{AD/AC}}} = sin(32°)
{{{6.69/AC}}} = sin(32°)
AC·sin(32°) = 6.69
AC = {{{6.69/sin("32°")}}}
AC = 12.62

Use the Pythagorean theorem to find BD

BD˛ + AD˛ = AB˛
      BD˛ = AB˛ - AD˛
      BD˛ = 10˛ - 6.69˛
      BD˛ = 55.2439
      BD = &#8730;<span style="text-decoration: overline">55.2439</span>
      BD = 7.43

Use the Pythagorean theorem to find DC

DC˛ + AD˛ = AC˛
      DC˛ = AC˛ - AD˛
      DC˛ = (12.62)˛ - (6.69)˛
      DC˛ = 114.5083
       DC = &#8730;<span style="text-decoration: overline">114.5083</span>
       DC = 10.70

We have side AB = 10, side AC = 12.63, and since we have
BD = 7.43 and DC = 10.70, we have side BC = BD+DC = 7.43+10.70 = 18.13

So we can find the perimeter AB + BC + AC = 10 + 18.13 + 12.62 = 40.75

Edwin</pre>