Question 1182028
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In ∆ACS, AC = 12, SA = 8 and SC = 16. Determine the measures of all the interior angles of the triangle.
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Use the cosine law

    c^2 = a^2 + b^2 - 2bccos(C),

where "a", "b", "c" are the lengths of the three sides of a triangle, and C is the angle opposite to the side "c".


It gives

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


Apply this formula to each angle of the triangle


    angle A, a= 12, b= 8, c= 16;  cos(A) = {{{(12^2 + 8^2 - 16^2)/(2*12*8)}}} = -0.25;

                                  A = arccos(0.25) = 1.823 radians = 104.48 degrees.


    angle C, a= 12, b= 16, c= 8;  cos(A) = {{{(12^2 + 16^2 - 8^2)/(2*12*16)}}} = 0.875;

                                  C = arccos(0.875) = 0.505 radians = 28.96 degrees.


    angle S, a= 8, b= 16, c= 12;  cos(S) = {{{(16^2 + 8^2 - 12^2)/(2*16*8)}}} = 0.6875;

                                  C = arccos(68.75) = 0.813 radians = 46.57 degrees.


The problem is just solved: all the angles are found.


As a conclusion, I will check the sum of angles  104.48 + 28.96 + 46.57 = 180 degrees,   ! correct !
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Solved.