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Ex 6.3, 2 - Chapter 6 Class 10 Triangles (Term 1)
Last updated at May 29, 2018 by Teachoo
Transcript
Ex 6.3, 2 In figure, ΔODC ∼ ΔOBA, ∠BOC = 125° and ∠CDO = 70°. Find ∠DOC, ∠DCO and ∠OAB. Given: Δ ODC ∼ Δ OBA ∠ BOC = 125° ∠ CDO = 70° To find: ∠DOC, ∠DCO and ∠OAB Solution: Here, BD is a line, So, we can apply linear pair on it. ∠BOC + ∠DOC = 180° 125° + ∠DOC = 180° ∠DOC = 180° – 125° ∠DOC = 180° – 125° ∠DOC = 55° Now in Δ DCO ∠CDO + ∠DCO + ∠DOC = 180° 70° + ∠DCO + 55° = 180° 125° + ∠DCO = 180° ∠DCO = 180° - 125° ∠DCO = 55° Now it is given that ΔODC ∼ ΔOBA Hence ∠DCO = ∠OAB 55° = ∠ OAB ∠OAB = 55°
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Ex 6.3
Ex 6.3, 1 (i)
Ex 6.3, 1 (ii)
Ex 6.3, 1 (iii)
Ex 6.3, 1 (iv)
Ex 6.3, 1 (v) Important
Ex 6.3, 1 (vi)
Ex 6.3, 2 You are here
Ex 6.3, 3
Ex 6.3, 4 Important
Ex 6.3, 5
Ex 6.3, 6
Ex 6.3, 7
Ex 6.3, 8 Important
Ex 6.3, 9
Ex 6.3, 10
Ex 6.3, 11 Important
Ex 6.3, 12 Important
Ex 6.3, 13 Important
Ex 6.3, 14 Important
Ex 6.3, 15 Important
Ex 6.3, 16
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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.