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AB is the chord of a circle with centre O and DOC is a line segment originating from a point D on the circle and intersecting, AB produced at C such that BC = OD. If ∠BCD = 20°, then ∠AOD =? (SSC CGL 2nd Sit. 2013)
- a)20°
- b)30°
- c)40°
- d)60°
Correct answer is option 'D'. Can you explain this answer?
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Most Upvoted Answer
AB is the chord of a circle with centre O and DOC is a line segment or...
Here BC = OD = radius {given}
In ΔBOC,
BC = OB (radius).
∴ ∠BOC = ∠BCO = 20°
∠OBA = ∠BOC + ∠BCO = 20° + 20° = 40°
Again, In DAOB,
AO = BO = radius
∠OAB = ∠OBA = 40°
∴ ∠AOB = 180° – 40° – 40° = 100°
Again,
∠AOD + ∠AOB + ∠BOC = 180°
∠AOD + 100° + 20° = 180°
∠AOD = 180° – 120° = 60°
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AB is the chord of a circle with centre O and DOC is a line segment or...
BC = OD, then triangle BCO is an isosceles triangle.
Since AB is a chord of the circle with center O, then angle OBA is equal to angle OAB.
Since triangle BCO is isosceles, then angle BOC is equal to angle BCO.
Therefore, angle OBA + angle BOC = angle OAB + angle BCO.
But angle OAB + angle BCO is equal to angle DOC, since DOC is a straight line.
Therefore, angle OBA + angle BOC = angle DOC.
Since angle OBA + angle BOC is equal to angle DOC, and angle OBA is equal to angle OAB, then angle OAB + angle OBA + angle BOC = angle DOC + angle OBA.
Since angle OAB + angle OBA + angle BOC = angle DOC + angle OBA, then angle OAB + angle BOC = angle DOC.
Therefore, triangle OAB is similar to triangle DOC by the Angle-Angle Similarity Postulate.
Since AB is a chord of the circle with center O, then angle OBA is equal to angle OAB.
Since triangle BCO is isosceles, then angle BOC is equal to angle BCO.
Therefore, angle OBA + angle BOC = angle OAB + angle BCO.
But angle OAB + angle BCO is equal to angle DOC, since DOC is a straight line.
Therefore, angle OBA + angle BOC = angle DOC.
Since angle OBA + angle BOC is equal to angle DOC, and angle OBA is equal to angle OAB, then angle OAB + angle OBA + angle BOC = angle DOC + angle OBA.
Since angle OAB + angle OBA + angle BOC = angle DOC + angle OBA, then angle OAB + angle BOC = angle DOC.
Therefore, triangle OAB is similar to triangle DOC by the Angle-Angle Similarity Postulate.
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AB is the chord of a circle with centre O and DOC is a line segment originating from a point D on the circle and intersecting, AB produced at C such that BC = OD. If ∠BCD = 20°, then ∠AOD =? (SSC CGL 2nd Sit. 2013)a)20°b)30°c)40°d)60°Correct answer is option 'D'. Can you explain this answer?
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