Given a triangle with a 9b 11 a 31degree what is are th

Given a triangle with a = 9,b = 11 \' a = 31degree , what is (are) the possible length(s) of c a. 14.21 b. 16.42 or3.41 c. 16.42 or2.44 d. 6.61

Solution

a = 9 , b = 11 , = 31^o

a/sin = b/sin

==> sin = (b/a)sin

==> sin = (11/9)sin(31^o)

==> sin = 0.6295

==> = 39.0126^o , 180^o - 39.0126^o         since sine function is positive in I and II Quadrants

==> = 39.013^o , 140.987^o

Sum of angles in a triangle = 180^o

==> + + = 180^o

==> = 180^o - (+ )

for = 39.013^o

==> = 180^o - (31^o + 39.013^o )

==> = 109.987^o

for = 140.987^o

==> = 180^o - (31^o + 140.987^o )

==> = 8.013^o

using cosine formula c² = a² + b² -2abcos

for = 109.987^o

==> c² = 9² + 11² -2(9)(11)cos109.987^o

==> c² = 269.68

==> c = 16.42

for = 8.013^o

==> c² = 9² + 11² -2(9)(11)cos8.013^o

==> c² = 5.933

==> c = 2.435 = 2.45

Hence c = 16.42 or 2.44

Hence option (c) is correct.