A,B,C and D are four points on a circle.AC and BD intersect at point E such that angle BEC=130 degree and angle ECD=20degree.Find angle BAC
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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.
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