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°