The sum of angles on a straight line equal 180^o. This will help us to find any angles on a straight line.

Ex: a) in the figure below, PQR is a straight line with QS = QR and TQS = 3y. Find TPQ

                         T                                                        S

                                3y^o                

                                                       3y^o

                  y^o                         x^o          56^o   

       P                                             Q                                                           R

Solution: since the sum of angles on a straight line is 180^o

x + 3y + 56 = 180, x + 3y = 180 – 56 = 124

x + 3y + y = 180, x + 4y = 180

x + 3y = 124

x + 4y = 180

         0 – y = -56                         y = 56^o

Substituting y = 56, we have x + 3(56) = 124

x + 168 = 124 = -44