Question 1077715


 In triangle {{{ABC}}}, {{{a=15cm}}}, {{{b=21cm}}}, and {{{m}}} < {{{C=90}}} degree. 

use Law of Cosines:

{{{a^2=b^2+c^2-2bc*cos(A)}}}
{{{b^2=a^2+c^2-2ac*cos(B)}}}
{{{c^2=a^2+b^2-2ab*cos(C)}}}

{{{c^2=15^2+21^2-2*15*21*cos(90)}}}......{{{cos(90)=0}}}
{{{c^2=225+441-630*0}}}
{{{c^2=666}}}
{{{c=sqrt(666)}}}
{{{c=25.8cm}}}


{{{a^2=b^2+c^2-2bc*cos(A)}}}
{{{a^2-b^2-c^2=-2bc*cos(A)}}}
{{{cos(A)=(b^2+c^2-a^2)/2bc}}}
{{{cos(A)=(21^2+(25.8)^2-15^2)/(2*21*25.8)}}}
{{{cos(A)=(441+665.64-225)/(1083.6)}}}
{{{cos(A)=(881.64)/(1083.6)}}}
{{{cos(A)=0.8136212624584718}}}

{{{A=cos^-1(0.8136212624584718)}}}

{{{A=35.55°}}}


{{{b^2-a^2-c^2=-2ac*cos(B)}}}

{{{2ac*cos(B)=a^2+c^2-b^2}}}

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

{{{cos(B)=(15^2+(25.8)^2-21^2)/(2*15*25.8)}}}

{{{cos(B)=(449.64)/(774)}}}

{{{cos(B)=0.580930232558}}}

{{{B=cos^-1(0.580930232558)}}}

{{{B=54.48}}}°