In Fig. 10.39, A, B, C and D are four points on a
circle. AC and BD intersect at a point E such
that Z BEC = 130° and Z ECD = 20°. Find
ZBAC.
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Given :
∠BEC = 130°
∠ECD = 20°
To find :
∠BAC
Solution :
Since AC is a straight line
∠AEB + ∠BEC = 180°
∠AEB + 130° = 180° [ linear pair ]
∠AEB = 180° - 130°
∠AEB = 50°
∠DEC = ∠AEB = 50° [ vertically opp. angles]
In ∆DEC,
∠DEC + ∠ECD + ∠BDC = 180° [ angle sum property ]
50° + 20° + ∠BDC = 180°
70° + ∠BDC = 180°
∠BDC = 180° - 70°
∠BDC = 110°
Therefore, ∠BAC = ∠BDC = 110° [ anglea in the same segment are equal ]
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