In Fig. 6.35, ∆ ODC ~ ∆ OBA, ∠ BOC = 125° and ∠ CDO = 70°. Find ∠ DOC, ∠ DCO and ∠ OAB
Solution:
We know that, if two angles of one triangle are respectively equal to two corresponding angles of another triangle, then the two triangles are similar.
This is referred to as AA criterion of similarity 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 of a triangle]
∠DCO = 180° - (55° + 70°)
∠DCO = 180° - 125°
∠DCO = 55°
In ΔODC and ΔOBA
ΔODC ∼ ΔOBA (given)
⇒ ∠DCO = ∠OAB
Thus, ∠OAB = 55°
☛ Check: NCERT Solutions Class 10 Maths Chapter 6
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
Summary:
In the figure ∆ ODC ~ ∆ OBA, ∠ BOC = 125° and ∠ CDO = 70°. The values of ∠ DOC, ∠ DCO and ∠ OAB are 55° each.
☛ Related Questions:
- State which pairs of triangles in Fig. 6.34 are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form:
- In Fig. 6.36 QR/QS = QT/PR and ∠1 = ∠2. Show that ΔPQS ~ ΔTQR.
- S and T are points on sides PR and QR of ∆PQR such that ∠P = ∠RTS. Show that ΔRPQ ~ ΔRTS.
- In Figure 6.37, if ∆ABE ≅ ∆ACD, show that ∆ADE ~ ∆ABC.