Question 802642
Triangles that are not right triangles can be solved by
Law of Sines (if you have or can get one side and the opposite angle)
or
Law of Cosines (otherwise).
 
Law of Cosines:
{{{drawing(300,300,-1,12,-1,12,
triangle(0,0,9,0,1.67,6.8),
locate(-0.3,-0.05,A),locate(4,0,c),
locate(9.3,-0.05,B),locate(1.4,7.5,C),
locate(5,4.4,a),locate(0.3,3.4,b)
)}}} {{{a^2=b^2+c^2=2ab*cos(A)}}}
The law of cosines can be used to find angle A when we just have the lengths of the 3 sides.
The largest angle is the one opposite the longest side, so in a triangle with side lengths 7,9,and 10, the largest angle will be opposite the side with length 10.
So we set {{{a=10}}} and try to find {{{A}}}.
{{{10^2=7^2+9^2-2*7*9*cos(A)}}}
{{{100=49+81-126cos(A)}}}
{{{100=130-126cos(A)}}}
{{{126cos(A)=130-100}}}
{{{126cos(A)=30}}}
{{{cos(A)=30/126=about0.2381}}} --> {{{A=76.2^o}}}