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°