Question

5.
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 < BEC = 130° and Z ECD = 20°. Find
ZBAC.​

Answer 1

Answer:

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Answer 2

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 ]