Question

In figure angle ADC is 130°.
Chord BC = BE.
Find angle cbe​

Answer 1

Answer:

If O is centre

then angle b is 90°

If not then....

Other Solution

In △BCO and △BEO

⇒ BC=BE [ Given ]

⇒ ∠BCO=∠BEO [ Base angles of equal

sides are also equal]

⇒ BO=BO [ Common side ]

∴ △BCO≅△BEO [ By SAS congruence

Rule]

⇒ ∠CBO=∠OBE [ C.P.C.T. ]

Quadrilateral ABCD is a cyclic quadrilateral since, its points lies on a circle.

⇒ ∠ADC+∠CBA=180° [ Sum of opposite angles of a cyclic quadrilateral is supplementary ]

⇒ 130°+∠CBA=180°

⇒ ∠CBA=50°

i.e. ∠CBO=50°

∴ ∠OBE=50°

⇒ ∠CBE=50°+50°

=100°

Answer 2

the whole part of the circle is 360° therefore

ADC +CBE=360°

130°+ x =360°

x =360°-130°

=270°