# In Fig. 6.35, ∆ ODC ~ ∆ OBA, ∠ BOC = 125° and ∠ CDO = 70°. Find ∠ DOC, ∠ DCO and ∠ OAB

**Solution:**

As we are aware if two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.

This is referred to as AA criterion for two triangles.

In the given figure.

∠DOC = 180 - ∠COB [∵ ∠DOC and ∠COB from a linear pair]

∠DOC = 180 - 125

∠DOC = 55

In ΔODC

∠DCO = 180 - (∠DOC + ∠ODC) [ angle sum property]

∠DCO = 180

∠DCO = 55

In ΔODC and ΔOBA

DODC ∼ DOBA

⇒ ∠DCO = ∠OAB

∠OAB = 55

**Video Solution:**

## In Fig. 6.35, ∆ ODC ~ ∆ OBA, ∠ BOC = 125° and ∠ CDO = 70°. Find ∠ DOC, ∠ DCO and ∠ OAB

### Class 10 Maths NCERT Solutions - Chapter 6 Exercise 6.3 Question 2:

In Fig. 6.35, ∆ ODC ~ ∆ OBA, ∠ BOC = 125° and ∠ CDO = 70°. Find ∠ DOC, ∠ DCO and ∠ OAB

In the figure above triangle ODC and OBA angle, BOC = 125 degrees and angle CDO = 70 degrees. Hence, angles DOC, DCO, and OAB are 55 degrees each.