Question 1177730
{{{D}}} = projection of {{{A}}} on  {{{BC}}}.

Then:

{{{b*sin(C) = AD}}} => height of the triangle {{{ABC}}} with respect to base{{{ BC}}}

{{{b*cos(C) = DC}}} =>projection of side {{{b = AC}}} onto base {{{BC}}}

{{{c*cos(B) = BD}}}=>projection of side {{{c = AB}}} onto base {{{BC}}}

Thus 

{{{b*cos(C)+c*cos(B) = BD + DC = BC = base}}}


So {{{42 = (b*sin(C))(b*cos(C)+c*cos(B)) = AD * BC = height*base = 2*((1/2)height*base) = 2(ABC)}}}

Therefore the area of {{{ABC = 42/2 = 21 }}} square units.