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Ex 6.3
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
Last updated at May 29, 2018 by Teachoo
Maths Crash Course - Live lectures + all videos + Real time Doubt solving!
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°