Question 255619
A triangle with a perimeter of 20cm. Two angles are 30 and 50. What are the lengths of the sides in cm.
<pre><font size = 4 color = "indigo"><b>
Since the sum of the three angles of any triangle is 180°, we
have that the third angle is 180° - (30°+50°) = 180° - 80° = 100° 

I'll arbitrarily let angle A = 30°, angle B = 100°, and angle C = 50°
{{{drawing(400,400,-.1,1.9,-.5,1.5,

triangle(0,0,cos(30*pi/180),0,cos(30*pi/180)+cos(50*pi/180),sin(100*pi/180)),
locate(.15,.1,"30°"), locate(1.25,.8,"50°"),
locate(0,0,A),locate(cos(30*pi/180),0,B), locate(.75,.1,"100°"), 
locate(cos(30*pi/180)+cos(50*pi/180),sin(100*pi/180),C),

locate(.5,0,c), locate(1.1,.35,a), locate(.7,.6,b)
)}}}
By the law of sines,

{{{a/sin("30°")=b/sin("100°")}}}

{{{a*sin("100°")=b*sin("30°")}}}

{{{a=(b*sin("30°"))/sin("100°")}}}


{{{c/sin("50°")=b/sin("100°")}}}

{{{c*sin("100°")=b*sin("50°")}}}

{{{c=(b*sin("50°"))/sin("100°")}}}

Perimeter = a + b + c = 20, so substituting in

{{{a + b + c = 20}}},

{{{(b*sin("30°"))/sin("100°")+b+(b*sin("50°"))/sin("100°")=20}}}

{{{b*sin("30°")+b*sin("100°")+b*sin("50°")=20*sin("100°")}}}

{{{b(sin("30°")+sin("100°")+sin("50°"))=20*sin("100°")}}}

{{{b= 20*sin("100°")/(sin("30°")+sin("100°")+sin("50°"))=8.750532397cm}}}

{{{a=(b*sin("30°"))/sin("100°")}}}

{{{a=((20*sin("100°"))/(sin("30°")+sin("100°")+sin("50°")))*((sin("30°"))/sin("100°"))}}}


{{{a=((20*cross(sin("100°")))/(sin("30°")+sin("100°")+sin("50°")))*((sin("30°"))/cross(sin("100°")))}}}

{{{a= 20*sin("30°")/(sin("30°")+sin("100°")+sin("50°"))=4.442761731cm}}}

-----

{{{c=(b*sin("50°"))/sin("100°")}}}

{{{c=(b*sin("50°"))/sin("100°")}}}

{{{c=((20*sin("100°"))/(sin("30°")+sin("100°")+sin("50°")))*((sin("50°"))/sin("100°"))}}}


{{{c=((20*cross(sin("100°")))/(sin("30°")+sin("100°")+sin("50°")))*((sin("50°"))/cross(sin("100°")))}}}

{{{c= 20*sin("50°")/(sin("30°")+sin("100°")+sin("50°"))}}}


{{{c= 20*sin("50°")/(sin("30°")+sin("100°")+sin("50°"))=6.806705873cm}}}


Edwin</pre>