Question 1204422


if
{{{alpha = 119}}}°,
{{{a = 13}}},
{{{b = 27}}}


since {{{alpha > 90}}}°, use the Law of Cosines equation:


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


{{{13^2=27^2+c^2-2*27c*cos(119)}}}

{{{169=729+c^2-54c*cos(119)}}}

{{{c^2 + 54c*sin((29pi)/180) + 560 = 0}}}

{{{c^2 + 26.179c+560=0}}}


using quadratic formula, we get following solutions:

{{{c = -13.0895 - 19.7146 *i}}}

or

{{{c = -13.0895+ 19.7146* i}}}


so, there is no real solution for side {{{c}}}

For {{{ASS}}} ({{{SSA}}}) theorem with {{{alpha >= 90}}}° ({{{alpha >= pi/2}}}) and {{{a <= b}}}, there are {{{no}}} solutions and {{{no}}}{{{ triangle}}}!


answer:  IMPOSSIBLE