Question 1079182
The numbers given for the interior angles of the triangle must be their measures in degrees.
They add to 180, as they should, and giving just two of them would be enough information.
Let it be
{{{x}}}= length of the shortest side, in cm
{{{y}}}= length of the medium length side, in cm
we know that in a triangle
shorter side is opposite smaller angle,
so the triangle looks like this:
{{{drawing(300,100,-17,7,-1.5,6.5,
triangle(-15.31,0,5.57,0,0,5.57),
arc(5.57,0,7,7,180,225),locate(2.4,1.8,45^o),
arc(-15.31,0,7,7,-20,0),locate(-11.7,1.8,20^o),
arc(0,5.57,4,4,45,160),locate(-3,5,115^o),
locate(-6,0,x+13cm),locate(2.8,3.8,x),
locate(-7.3,4.4,y)
)}}}
Law of sines agrees that shorter side is opposite smaller angle,
but goes further and says that
the ratio of side length to sine of opposite angle is the same for all sides.
So {{{x/sin(20^o)=(x+13)/sin(115^o)=y/sin(45^o)}}}
we can find the approximate values of those sines.
{{{sin(20^o)=0.34202}}} , {{{sin(115^o)=0.90631}}} , and {{{sin(45^o)=0.70711}}} .
We can solve {{{x/sin(20^o)=(x+13)/sin(115^o)}}} for {{{x}}} .
{{{x/sin(20^o)=(x+13)/sin(115^o)}}}
{{{x/0.34202=(x+13)/0.90631}}}
{{{0.90631x=0.34202(x+13)}}}
{{{0.90631x=0.34202x+0.34202*13}}}
{{{0.90631x-0.34202x=0.34202*13}}}
{{{(0.90631-0.34202)x=0.34202*13}}}
{{{0.56429x=0.34202*13}}}
{{{0.56429x=4.44626}}}
{{{x=4.44626/0.56429}}}
{{{x=approximately 7.88}}}
So, {{{x+13=approximately}}}{{{7.88+13=20.88}}}
From {{{x/sin(20^o)=y/sin(45^o)}}} we find {{{y}}} :
{{{x/sin(20^o)=y/sin(45^o)}}}
{{{x/0.34202=y/0.70711}}}
From {{{0.56429x=0.34202*13}}} (above), we know that {{{x/0.34202=13/0.56429}}} ,
so we can use that rather than the already rounded {{{x=approximately 7.88}}} .
{{{y/0.70711=13/0.56429}}}
{{{y=13*0.70711/0.56429}}}
{{{y=approximately 16.29}}}
S]Now, knowing the lengths of all 3 sides to 2 decimal places,
we add them to find the approximate perimeter,
to be reported further rounded as a whole number.
{{{20.88cm+16.29cm+5.88cm=45.05cm}}}
To the nearest cm, the perimeter is {{{highlight(45cm)}}} .