Question

In the given figure, if ZBEC = 135'. ZDCE = 25°, then find ZBAC.​

Answer 1

Answer:

∠BAC = ∠CDE (Angles in the segment of the circle)

In ΔCDE,

∠CEB = ∠CDE + ∠DCE (Exterior angles of the triangle)

⇒ 120° = ∠CDE + 25°

⇒ ∠CDE = 95°

Thus, ∠BAC = 95°

∠BEC is exterior angle of △CDE

∴∠CDE + ∠DCE = ∠BEC

⇒∠CDE + 25° = 120°

⇒ ∠CDE = 95°

Now, ∠BAC = ∠CDE [∵ angle in same segment are equal]