In Fig. 6.35, ΔODC ~ ΔOBA, BOC = 125° and CDO = 70°. Find DOC, DCO and OAB


From the figure,



We see, DOB is a straight line


DOC + COB = 180°  (angles form a supplementary pair)


DOC = 180° - 125°


∠ DOC= 55°


Now, In ΔDOC,


DCO + CDO + DOC = 180°


(Sum of the measures of the angles of a triangle is 180°)


DCO + 70° + 55° = 180°


DCO = 55°


It is given that ΔODC ΔOBA


OAB = OCD (Corresponding angles are equal in similar triangles)


Thus, ∠ OAB = 55°

7