Find angle CBE in question 15
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We have, ∠ADC = 130° and chord BC – chord BE. Suppose, we consider the points A, B, C and D form a cyclic quadrilateral.
Since, the sum of opposite angles of a cyclic quadrilateral ΔDCB is 180°.
∠ADC + ∠OBC = 180°
⇒ 130°+ ∠OBC = 180°
⇒ ∠OBC = 180° -130° = 50°
In ΔBOC and ΔBOE,
BC = BE [given equal chord]
OC =OE [both are the radius of the circle]
and OB = OB [common side]
ΔBOC ≅ΔBOE
∠OBC = ∠OBE = 50° [by CPCT]
∠CBE = ∠CBO + ∠EBO = 50° + 50° = 100°
We have, ∠ADC = 130° and chord BC – chord BE. Suppose, we consider the points A, B, C and D form a cyclic quadrilateral.
Since, the sum of opposite angles of a cyclic quadrilateral ΔDCB is 180°.
∠ADC + ∠OBC = 180°
⇒ 130°+ ∠OBC = 180°
⇒ ∠OBC = 180° -130° = 50°
In ΔBOC and ΔBOE,
BC = BE [given equal chord]
OC =OE [both are the radius of the circle]
and OB = OB [common side]
ΔBOC ≅ΔBOE
∠OBC = ∠OBE = 50° [by CPCT]
∠CBE = ∠CBO + ∠EBO = 50° + 50° = 100°
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