Question 1192620


given:  {{{(b*sin( C)) (b*cos (C) + c*cos( B)) =42}}}


{{{b*cos(C) + c*cos(B)}}} is the length of the side {{{a}}} of the triangle ABC


use the law of cosine to prove it:


{{{cos(C) =  (a^2 + b^2 - c^2)/(2ab)}}}

{{{cos(B) =  (c^2 + a^2 - b^2)/(2ca)}}}


then we have

{{{b*cos (C) + c*cos( B)}}}

={{{b((a^2 + b^2 - c^2)/(2ab))+c( (c^2 + a^2 - b^2)/(2ca))}}}

={{{((a^2 + b^2 - c^2)/(2a))+( (c^2 + a^2 - b^2)/(2a))}}}

={{{(a^2 + b^2 - c^2+c^2 + a^2 - b^2)/(2a)}}}

={{{(2a^2)/(2a)}}}

={{{a}}}


then we have

{{{b*sin(C)}}} is the length of the altitude drawn from vertex A to the side {{{a}}}, and the area of triangle is:

{{{area=(1/2)ab*sin( C) }}}

 {{{area=(1/2)42}}}

{{{area=21}}} sq. units