Question 1076294
Using the law of cosines,
{{{c^2=a^2+b^2-2ab*cos(C)}}}
{{{cos(C)=(a^2+b^2-c^2)/(2ab))}}} 
So,
{{{cos(C)=(3^2+5^2-7^2)/(2(3)(5))}}} 
{{{cos(C)=(9+25-49)/30}}} 
{{{cos(C)=(-15)/30}}} 
{{{cos(C)=-1/2}}} 
{{{C=120}}} 
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To find the area, use Heron's formula,
{{{S=(3+5+7)/2=15/2}}}
So then,
{{{A=sqrt(S(S-3)(S-5)(S-7))}}}
{{{A=sqrt((15/2)(15/2-6/2)(15/2-10/2)(15/2-14/2))}}}
{{{A=sqrt(675/16)}}}
{{{A=(15/4)sqrt(3)}}}