InterviewSolution
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In the figure, ∠BAD = 40° then find ∠BCD. |
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Answer» ‘O’ is the centre of the circle. ∴ In ΔOAB; OA = OB (radii) ∴ ∠OAB = ∠OBA = 40° (∵ angles opp. to equal sides) Now ∠AOB = 180° – (40° + 40°) (∵ angle sum property of ΔOAB) = 180°-80° = 100° But ∠AOB = ∠COD = 100° Also ∠OCD = ∠ODC [OC = OD] = 40° as in ΔOAB ∴ ∠BCD = 40° (OR) In ΔOAB and ΔOCD OA = OD (radii) OB = OC (radii) ∠AOB = ∠COD (vertically opp. angles) ∴ ΔOAB ≅ ΔOCD ∴ ∠BCD = ∠OBA = 40° [ ∵ OB = OA ⇒ ∠DAB = ∠DBA] |
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