Question

In the given figure , ∠ ABC = 130 ° and chord BC = chord BE. Find ∠CBE.​

Answer 1

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In the given figure, ∠ADC =130° and chord BC = chord BE

The point A, B, C and D formed a cyclic quadrilateral .

∴ ∠ADC + ∠OBC = 180°

⇒ 130°+ ∠OBC = 180°

⇒ ∠OBC = 180°

Now, in ΔBOC and ΔBOE ,

BC = BE [given]

OC = OE [radii of the same circle]

OB = OB [common side]

∴ ΔBOC ≅ ΔBOE [by SSS congruent rule]

Then , ∠OBC = ∠OBE = 50° [CPCT]

∴ ∠CBE = ∠CBO + ∠EBO = 50° + 50° = 100°

Answer 2

Step-by-step explanation:

Given

< ADC = 130°

Chord BC = Chord BE

Find < CBE

Solutions

Quadrilateral ABCD is Cyclic Quadrilateral

Therefore End Point In Circumference Of Circle

The Sum Of Opposite Angle Of Cyclic Quadrilateral is 180°

So , < ADC + < ABC = 180°

< ABC = 180° - < ADC

< ABC = 180° - 130°

< ABC = 50°. ( i )

Now in ∆ OBC and ∆ OBE

OB = OB ( Common )

BC = BE ( Given )

OC = OE ( Radius of Circle )

By SSS Rule

∆ OBC Congruent ∆ OBE

So , < OBC = < OBE ( C. P. C . T )

< OBC = < OBE = 50° From ( i )

< CBE = < OBC + < OBE

< CBE = 50° + 50°

< CBE = 100°

The < CBE is 100°