Question 656637
Knowing the lengths, {{{a}}} and {{{b}}}, of two sides of a triangle,
and the measure, {{{C}}} of the angle those sides make,
you can calculate the area of the triangle as
{{{Area=a*b*sin(C)}}}
{{{drawing(300,300,-2,16,-2,16,
triangle(0,0,12,0,6.81,13.37),
locate(-0.5,-0.1,C),locate(12.1,-0.1,A),
locate(6.9,14,B), locate(5.8,-0.1,b),
locate(2.8,7.2,a),locate(6.9,3.5,"h=a*sinC"),
green(line(6.81,0,6.81,13.37))
)}}} or {{{drawing(300,300,-2,16,-2,16,
triangle(0,0,15,0,5.45,10.69),
locate(-0.5,-0.1,C),locate(15.1,-0.1,B),
locate(5.5,11.5,A), locate(7,-0.1,a),
locate(2.8,7.2,b),locate(5.5,3.5,"h=b*sinC"),
green(line(5.45,0,5.45,10.69))
)}}} 
{{{Area=a*b*sin(C)}}}
{{{a=15}}} , {{{b=12}}}, {{{C=63^o}}}
{{{Area=15*12*sin(63^o)}}}
{{{sin(63^o)=approx.}}}{{{0.891}}} (rounding)
{{{Area=approx.}}}{{{15*12*0.891}}}
{{{Area=approx.}}}{{{highlight(160)}}}