Question 999906
Triangle ABC has side {{{a = 11}}},{{{ b = 14}}} and {{{c = 17}}}.

use The Law of Cosines (also called the Cosine Rule)


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

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

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

{{{cos(A)=(17^2+14^2-11^2 )/(2*14*17)}}}

{{{cos(A)=(289+169-121 )/476}}}

{{{cos(A)=(337 )/476}}}

{{{cos(A)=0.7079831932773109}}}

{{{A=cos^(-1)(0.7079831932773109)}}}

{{{A=40.1192}}}°

{{{A=40}}}°