Question

29) In the fig. ZABC = 69°, ZACB = 31° find (a) ZBDC b) ZBEC
A
D
69° 31°
B.
C
E
301 The​

Answer 1

In ∆ABC,

∠BAC + ∠ABC + ∠ACB = 180° (Sum of all the angles of a triangle)

∠BAC + 69° + 31° = 180°

⇒ ∠BAC = 180° – 100°

⇒ ∠BAC = 80°

(a) ∴ ∠BDC = ∠BAC = 80° (Angles in the same segment of a circle are equal)

Now, you can easily find ∠BEC. So, do it yourself...!!