Question

In the given figure A,B,C and D are four points on a circle AC and BD intersect at E such that angle BEC= 130 degree and Angle ECB =20 degree .Find angle BAC.​

Answer 1

Answer:  angle ABE = angle DEC (vertically opposite)

              angle ECD = angle EBA

      angle aed = 130 (vertically opposite)

130 + 130 + 2x = 360

2x = 260

x =130

angle AEB = 130

angle ABE = 20

20 + 130 + angle BAC = 180 (angle sum property)

150 + angle BAC  = 180

angle BAC = 30

Step-by-step explanation

Answer 2

Hello mate =_=

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Solution:

∠BEC+∠DEC=180°                (Linear Pair)

⇒∠130°+∠DEC=180°

⇒∠DEC=180°−130°=50°             ......(1)

∠DEC+∠DCE+∠BDC=180°

Putting value of ∠DCE and using (1), we get

50°+20°+∠BDC=180°

⇒∠BDC=180°−50°−20°=110°    

We also have ∠BDC=∠BAC  (Angles in the same segment are equal.)

⇒∠BAC=110°

hope, this will help you.

Thank you______❤

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